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AN 


ELEMENTARY    TREATISE 


MECHANICS, 


EMBRACING    THE 


THEORY  OF  STATICS  AND  DYNAMICS, 


AND    ITS    APPLICATION    TO 


SOLIDS  AND  FLUIDS. 


^veparcij  for  tlje  £intier=£va*iuate  ©ouvsc  in  tije  S&cskjjau  iHmbcvsftg. 


BY  AUGUSTUS  W.  SMITH,  LL.D., 


NEW    YORK: 

HARPER  &  BROTHERS,  PUBLISHERS. 

18  68. 


1°[ 


*3l  I 
S 


Entered,  according  to  Act  of  Congress,  in  the  year  one  thousand 
eight  hundred  and  forty-nine,  by 

Harper  &  Brothers, 

ia  the  Clerk's  Office  of  the  District  Court  of  the  Southern  District 
of  New  York. 


PREFACE, 


The  preparation  of  the  present  treatise  was  undertaken  un- 
der the  impression  that  an  elementary  work  on  analytical  Me- 
chanics, suited  to  the  purposes  and  exigencies  of  the  course  of 
study  in  colleges,  was  needed.  This  impression  is  the  result 
of  long  experience  in  teaching,  and  a  fair  trial  of  all  known 
American  works  and  reprints  designed  for  such  use. 

It  can  scarcely  be  necessary,  at  this  period,  to  assign  at 
length  the  reasons  for  adopting  the  analytical  methods  of  in- 
vestigation. Whether  the  object  be  intellectual  discipline,  or 
a  knowledge  of  facts  and  principles,  or  both,  the  preference 
must  be  given  to  the  modern  analysis.  It  affords  a  wider  field 
for  the  exercise  of  judgment,  calls  more  fully  into  exercise  the 
inventive  powers,  and  taxes  the  memory  less  with  unimportant 
particulars,  thus  developing  and  strengthening  more  of  the 
mental  faculties,  and  more  equably  by  far  than  the  geometric- 
al methods.  It  is  more  universal  in  its  application,  shorter  in 
practice,  and  far  more  fruitful  in  results.  It  is,  indeed,  the  only 
method  by  which  the  student  can  advance  beyond  the  merest 
rudiments  of  the  science,  without  an  expense  of  time  and  en- 
ergy wholly  disproportioned  to  the  ends  accomplished — the 
only  method  by  which  he  can  acquire  a  self-sustaining  and  a 
progressive  power. 

As  the  hope  of  furnishing  to  the  student  some  additional 
facilities  for  a  pleasant  and  profitable  prosecution  of  this 
branch  of  study  was  the  motive  for  undertaking  the  prepara- 
tion of  this  manual,  it  will  be  proper,to  refer  to  some  of  the 

2 


PREFACE. 


specific  objects  had  in  view.  The  most  formidable  obstacles 
to  the  acquisition  of  any  branch  of  science  are  generally  found 
at  the  very  outset.  It  has  therefore  been  a  specific  object  to 
introduce  the  subject  by  giving  distinctness  to  the  elementary 
truths,  dwelling  upon  them  till  they  are  rendered  familiar, 
adopting  the  simplest  mode  of  investigation  and  proof  consist- 
ent with  rigor  of  demonstration,  and  avoiding  all  reference  to 
the  metaphysics  of  the  science  as  out  of  place  in  a  work  de- 
signed for  beginners.  At  every  successive  stage  of  advance- 
ment the  student  is  required  to  review  the  ground  passed  over 
by  the  use  of  the  principles  learned,  in  the  solution  of  exam- 
ples which  will  require  their  application,  and  test  at  once  his 
knowledge  of  them  and  his  ability  to  apply  them.  Having 
examined  in  this  way  each  division  of  forces  as  classified, 
more  general  methods  are  introduced,  and  their  application 
illustrated  by  numerous  problems.  These  methods  are  often 
employed  in  particular  cases  when  others  less  general  would 
be  shorter ;  for  it  is  a  readiness  in  their  use  and  a  familiarity 
with  their  application  that  gives  to  the  student  his  power  over 
more  difficult  and  complicated  questions. 

To  adapt  the  work  to  the  exigencies  of  the  recitation-room, 
the  whole  is  divided,  at  the  risk,  perhaps,  of  too  much  apparent 
formality,  into  distinct  portions  or  propositions,  suitable  for  an 
individual  exercise.  In  each  case  the  object  to  be  accomplish- 
ed is  distinctly  proposed,  or  the  truth  to  be  established  is 
briefly  and  clearly  enunciated.  The  student  has  thus  a  defin- 
ite object  before  him  when  called  upon  to  recite,  acquires  a 
convenient  formulary  of  words  by  which  to  quote  and  apply 
his  arguments,  and  the  more  clearly  conceives  and  marks  his 
own  progress. 

In  this  institution  the  mathematical  and  experimental  courses 
are  assigned  to  different  departments.  It  is  not,  however,  for 
this  reason  alone  that  I.have  purposely  abstained  from  swell- 


PREFACE.  ill 

ing  the  volume  with  diffuse  verbal  explanations,  and  the  intro- 
duction of  experimental  illustrations.  The  experienced  teach- 
er will  always  have  at  command  an  abundance  of  matter  of 
this  kind  to  meet  every  emergency,  and  can  adapt  the  kind 
and  mode  of  illustration  to  the  specific  difficulty  that  arises  in 
the  mind  of  his  pupil.  There  is  an  objection  to  the  introduc 
tion  of  such  matter,  growing  out  of  a  tendency,  in  some  at 
least,  to  become  the  passive  recipients  of  their  mental  aliment. 
These  mental  dyspeptics  loathe  that  which  costs  them  labor, 
and,  satisfied  with  the  stimulus  of  inflation,  seize  upon  the 
lighter  portions,  and  neglect  that  which  alone  can  impart 
vigor  to  their  mental  constitution.  Whoever  caters  for  this 
class  of  students  must  share  in  the  responsibility  of  raising  up 
a  race  effeminate  in  mind,  if  not  in  body. 

The  investigations  are  limited  to  forces  in  the  same  plane, 
except  in  the  case  of  parallel  forces.  This  limit,  while  it  is 
sufficiently  ample  to  include  the  most  important  topics  of  ter- 
restrial mechanics,  and  to  embrace  an  interesting  field  of  celes- 
tial mechanics,  is  fully  sufficient  to  occupy  the  time  which  the 
crowded  course  of  study  in  our  colleges  will  admit  of  being 
appropriated  to  it.  It  may,  without  marring  the  integrity  of 
the  work,  be  very  much  reduced  by  the  omission  of  all  those 
portions  in  which  the  integral  calculus  is  employed ;  and  still 
further,  if  desired,  by  omitting  all  that  relates  to  the  principle 
of  virtual  velocities,  and  its  application  to  the  mechanical 
powers.  A  copious  analysis  is  given  in  the  contents,  conven- 
ient for  frequent  and  rapid  reviews,  and  suggestive  of  ques- 
tions for  examination. 

In  this  work  no  claim  is  advanced  to  originality.  The  ma- 
terials have  been  sought  and  freely  taken  from  all  available 
sources.  Nearly  all  the  matter,  in  some  form,  is  found  in  al- 
most every  author  consulted,  and  credit  could  not  be  given  in 
every  case,  if,  in  an  elementary  work  designed  as  a  text-book, 


IV  PREFACE. 

it  were  desirable  to  do  so.  In  the  portion  on  Statics,  I  am 
most  indebted  to  the  excellent  introductory  work  of  Profess- 
or Potter,  of  University  College,  London.  The  chapter  on 
Couples  is  substantially  taken  from  Poinsot's  Elemens  de  Sta- 
tique.  The  questions,  generally  simplified  in  their,  character, 
are  mostly  taken  from  Walton's  Mechanical  and  Hydrostatical 
Problems,  and  Wrigley  and  Johnston's  Examples.  The  works 
more  especially  consulted  are  those  of  Poisson,  Francaeur, 
Gregory,  Whewell,  Walker,  Moseley,  and  Jamieson.  Some- 
thing has  been  taken  from  each,  but  modified  to  suit  the  spe- 
cific object  kept  constantly  in  view — the  preparation  of  a 
manual  which  should  be  simple  in  its  character,  would  most 
naturally,  easily,  and  successfully  induct  the  student  into  the 
elementary  principles  of  the  science,  and  prepare  him,  if  so 
disposed,  to  prosecute  the  study  further,  without  the  necessity 
of  beginning  again  and  studying  entirely  new  methods.  How 
far  I  have  succeeded  must  be  left  to  the  decision  of  others, 
especially  of  my  co-laborers  in  this  department  of  instruction. 

Wesleyan  University, 
Middletown,  Conn.,  Jan.,  1849. 


I^H1 


CONTENTS. 


Irt.  t*& 

INTRODUCTION 1 

1.  Definition  of  Mechanics  and  its  Subdivisions. 

2.  Definition  of  Force  and  its  Mechanical  Effects. 

3.  Definition  of  its  Intensity  and  Measure,  its  Direction  and  Point  of  Application. 

4.  Definition  of  Analytical  Mechanics. 

5.  Definition  of  Concurring  and  Conspiring  Forces. 

6.  Definition  of  Body,  Rigid,  Flexible,  and  Elastic. 


STATICS. 

CHAPTER  I. 

COMPOSITION   AND   EQUILIBRIUM    OF    CONCURRING   FORCES    IN    THE   SAME   PLANE.    3 

7.  Two  equal  and  opposite  Forces  in  Equilibrium. 

8.  Two  Forces  inclined  to  each  other  can  not  Equilibrate. 

9.  Definition  of  Resultant  and  Components. 

10.  Resultant  of  several  Conspiring  Forces. 

11.  Resultant  of  two  unequal  opposite  Forces. 
12    Resultant  of  any  number  of  opposite  Forces. 

13.  Point  of  Application  at  any  Point  in  its  Direction. 

14.  Direction  of  the  Resultant  of  several  Forces. 

15.  Direction  of  the  Resultant  of  two  equal  Forces. 
1C.  Direction  of  the  Resultant  of  two  unequal  Forces. 

17.  Variation  of  the  Magnitude  of  the  Resultant  and  its  Components. 

18.  Equilibrium  of  three  equal  Forces. 

19.  Resultant  of  two  equal  Forces  at  an  Angle  of  120°. 

20.  Each  of  three  equilibrating  Forces  equal  to  the  Resultant  of  the  other  two  in 

Magnitude. 

21.  Parallelogram  of  Forces. 

22.  Triangle  of  Forces. 

23.  Resolution  of  Forces. 

24.  Polygon  of  Forces. 

25.  Representation  of  equilibrated  Forces. 

26.  Graphical  determination  of  Resultant. 

27.  Parallelopiped  of  Forces. 

28.  Ratios  of  three  equilibrated  Forces. 

29.  Expression  for  the  Resultant  of  two  Forces. 

30.  Definition  of  Moment  of  a  Force — Origin  of  Moments. 

31.  Equality  of  the  Moment  of  the  Resultant  with  the  Sum  of  the  Moments  of  two 

Components. 
3-2.  Equality  of  the  Moment  of  the  Resultant  with  the  Sum  of  the  Momenta  ol  any 
member  of  Components. 

33.  When  the  Origin  of  Moments  is  fixed. 

34.  When  the  Forces  are  in  Equilibrium. 

35.  Examples. 


VI  CONTENTS. 

Art.  Pag, 

CHAPTER  II. 

PARALLEL     FORCES 15 

'J6.  Resultant  of  two  Parallel  Forces. 

37.  Definition  of  Amis  of  Forces. 

38.  Equilibrium  of  two  Parallel  Forces  by  a  third  Force. 

39.  Point  of  Application  of  the  Resultant. 

40.  When  the  Forces  act  in  opposite  Directions. 

41.  When  the  Forces  are  Equal  and  Opposite. 

42.  Such  Forces  constitute  a  Statical  Couple. 

43.  Resultant  of  any  Number  of  Parallel  Forces. 

44.  Definition  of  Center  of  Parallel  Forces. 

45.  Equality  of  the  Moment  of  the  Resultant  with  the  Sum  of  the  Moments  of  the 

Components. 
40.  Definition  of  Moment  of  a  Force-  in  reference  to  a  Plane. 

47.  Conditions  of  Equilibrium  of  any  Number  of  Parallel  Forces. 

48.  Condition  of  Rotation. 

49.  When  in  Equilibrium,  each  Force  equal  in  Magnitude  to  the  Resultant  of  all  the 

others. 

50.  Equilibrium  independent  of  their  Direction. 

51.  Examples. 

CHAPTER  III. 

THEORY    OF    COUPLES 24 

52.  Definition  of  a  Statical  Couple. 

53.  Definition  of  the  Arms  of  a  Couple. 

54.  Definition  of  the  Moment  of  a  Couple. 

55.  A  Couple  may  be  turned  round  in  its  own  Plane. 

56.  A  Couple  may  be  removed  parallel  to  itself  in  its  own  Plane. 

57.  A  Couple  may  be  removed  to  a  Parallel  Plane. 

58.  Couples  are  equivalent  when  their  Planes  are  Parallel  and  Moments  are  EquaL 

59.  Couples  may  be  changed  into  others  having  Amis  of  a  given  Length. 
GO.  Definition  of  the  Axis  of  a  Couple. 

61.  Properties  of  an  Axis. 

62.  Definition  of  the  Resultant  of  two  or  more  Couples. 

63.  Equality  of  the  Moment  of  the  Resultant  with  the  Sum  of  the  Moments  of  the 

Components. 

64.  Equality  of  the  Axis  of  the  Resultant  with  the  Sum  of  the  Axes  of  the  Com- 

ponents. 

65.  The  Resultant  of  two  Couples  inclined  to  each  other. 

66.  Representative  of  the  Axis  of  the  Resultant  of  two  Couples. 

67.  Parallelogram  of  Couples. 

CHAPTER  IV. 

ANALYTICAL   STATICS  IN  TWO   DIMENSIONS 31 

68.  Resultant  of  any  Number  of  Concurring  Forces. 

69.  Directions  of  the  Rectangular  Components  involved  in  their  Trigonometrical  Values 

70.  Conditions  of  Equilibrium  of  Concurring  Forces. 

71.  Resultant  Force  and  Resultant  Couple  when  the  Forces  do  not  concur. 

72.  Construction  of  the  Results. 

73.  Equation  of  the  Resultant. 

74.  Equilibrium  of  non-concurring  Forces. 

75.  Equilibrium  when  there  is  a  fixed  Point  in  the  System, 

76.  Equilibrium  of  a  Point  on  a  Plane  Curve. 


CONTENTS.  Vll 

trt  Page 

77.  Conditions  of  Equilibrium. 

78.  Definition  of  Virtual  Velocitius. 

79.  Principle  of  Virtual  Velocities. 

80.  Principle  of  Virtual  Velocities  obtains  in  Concurring  Forces  in  the  same  Plane. 

,  81.  Principle  of  Virtual  Velocities  obtains  in  non-concurring  Forces  in  the  same  Plane. 
L»2.  The  Converse. 

CHAPTER  V. 

CENTER    OF     GRAVITY 43 

83.  Definition  of  Gravity. 

84.  Laws  of  Gravity. 

85.  Definition  of  a  Heavy  Body. 

66.  Definition  of  the  Weight  of  a  Body. 

67.  Definition  of  its  Mass. 

88.  Expression  for  Weight. 

89.  Definition  of  Density. 

00.  Another  Expression  for  Weight. 

91.  Relations  of  Masses  to  Volumes  of  the  same  Density. 

92.  Relations  of  Densities  to  Volumes  of  the  same  Mass. 

93.  Relations  of  Densities  to  Masses  of  the  same  Volume. 

94.  Definition  of  Center  of  Gravity. 

95.  Connection  of  the  Center  of  Gravity  with  the  Doctrine  of  Parallel  Forces. 

96.  Definition  of  a  Body  symmetrical  with  respect  to  a  Plane. 

97.  Position  of  its  Center  of  Gravity. 

98.  Definition  of  a  Body  symmetrical  with  respect  to  an  Axis. 

99.  Position  of  its  Center  of  Gravity. 

100.  Center  of  Gravity  of  a  Body  symmetrical  with  respect  to  two  Axes. 

101.  Definition  of  Center  of  Figure. 

102.  Center  of  Gravity  of  any  Number  of  heavy  Particles. 

103.  Their  Center  of  Gravity  when  their  Positions  are  given  by  their  Co-ordinates 

104.  Their  Center  of  Gravity  when  they  are  all  in  the  same  Line. 

105.  Their  Center  of  Gravity  when  they  are  Homogeneous. 

106.  The  Center  of  Gravity  of  the  Whole  and  a  Part  given  to  find  that  of  the  other  Part. 

107.  Examples — 1.  Of  a  Straight  Line — 2.  Triangle— 3.  Parallelogram — 4.  Polygon — 

5.  Triangular  Pyramid — 6.  Any  Pyramid — 7.  Frustum  of  a  Cone — 8.  Perimeter 
of  a  Triangle — 9.  Of  a  Triangle  in  Terms  of  its  Co-ordinates. 

CONDITIONS    OF   EQUILIBRIUM   OF   BODIES   FROM   THE   ACTION   OF  GRAVITY...    54 

108.  When  the  Body  has  a  Fixed  Point  in  it. 

109.  Definition  of  Stable,  Unstable,  and  Neutral  Equilibrium. 

110.  Position  of  the  Centerof  Gravity  when  the  Equilibrium  is  Stable  aud  when  Unstable. 

111.  Pressure  on  the  Fixed  Point. 

112.  Position  and  Pressure  when  there  are  two  Fixed  Points. 

113.  Position  and  Pressure  when  there  are  three  Fixed  Points. 

114.  Position  and  Pressure  when  a  Body  touches  a  Horizontal  Plane  in  one  Point. 

115.  Position  and  Pressure  when  a  Body  touches  a  Horizontal  Plane  in  two  Points. 

116.  Position  and  Pressure  when  a  Body  touches  a  Horizontal  Plane  in  three  Points 

117.  Position  and  Pressure  when  a  Body  touches  a  Horizontal  Plane  in  any  Number  of 

Points. 

118.  Measure  of  the  Stability  on  a  Horizontal  Plane. 

119.  Case  of  a  Body  on  an  Inclined  Plane. 

120.  Examples. 

APPLICATION  OF  THE  INTEGRAL  CALCULUS  TO  THE  DETERMINATION  OF  THE  CEN. 
TER  OF  GRAVITY 53 


Vlil  CONTENTS. 

Art  P'S* 

121.  General  differential  Expressions  for  the  Co-ordimites  of  the  Center  of  Gravity. 

122.  General  differential  Expressions  for  the  Center  of  Gravity  of  a  Plane  Curve. 

123.  General  differential  Expressions  for  the  Center  of  Gravity  of  a  Plane  Area. 

124.  General  differential  Expressions  for  the  Co-ordinates  of  a  Surface  of  Revolution. 

125.  General  differential  Expressions  for  the  Co-ordinates  of  a  Solid  of  Revolution. 

126.  Determination  of  a  Surface  of  Revolution. 

127.  Determination  of  a  Solid  of  Revolution. 

128.  Examples— 1.  Of  a  Circular  Arc— 2.  Circular  Segment— 3.  Surface  of  a  Spherical 

Segment — 4.  Spherical  Segment. 

129.  Examples  on  the  preceding  Chapters. 

CHAPTER  VI. 

THE   MECHANICAL   POWERS 7fe 

130.  Classification  of  the  Mechanical  Powers. 

§  I.   THE  LEVER. 

131.  Definition  of  a  Lever. 

132.  Kinds  of  Lever. 

133.  Conditions  of  Equilibrium  when  the  Forces  are  Parallel 

134.  Conditions  of  Equilibrium  when  the  Forces  are  Inclined. 

135.  Conditions  of  Equilibrium  when  the  Lever  is  Bent  or  Curved. 

136.  Conditions  of  Equilibrium  when  any  Number  of  Forces  in  the  same  Plane  act  on 

a  Lever  of  any  Form — Examples. 

§11.   WHEEL   AND    AXLE 86 

137.  Definition  of  Wheel  and  Axle. 

138.  Conditions  of  Equilibrium  when  two  Forces  act  Tangentially  to  the  Surface  of  the 

Wheel  and  Axle. 

139.  Perpetual  Lever. 

140.  Pressure  on  the  Axis. 

141.  Conditions  of  Equilibrium  of  any  Number  of  Forces. 

142.  Definition  of  Cogged  Wheels,  Crown  Wheels,  Beveled  Wheels,  Piuions,  Leaver. 

143.  Conditions  of  Equilibrium  in  Cogged  Wheels. 

144.  Conditions  of  Equilibrium  when  the  Cogs  are  of  equal  Breadth. 

145.  Conditions  of  Equilibrium  in  Cogged  Wheels  and  Pinions. 

§  III.  the  cord 91 

146.  Definition  of  the  Cord— Of  Tension. 

147.  Conditions  of  Equilibrium  when  there  are  three  Forces. 

148.  Conditions  of  Equilibrium  when  the  Ends  are  fixed. 

149.  Conditions  of  Equilibrium  when  the  Ends  are  fixed  and  a  third  Force  applied  to 

a  Running  Knot. 

150.  Conditions  of  Equilibrium  when  there  is  any  number  of  Forces. 

151.  Relations  of  the  Forces  when  in  Equilibrium. 

152.  Relations  of  the  Forces  when  the  Ends  are  Fixed  and  the  Forces  Parallel. 

153.  Relations  of  the  Forces  when  the  Ends  are  Fixed  and  the  Forces  are  Weights. 

154.  Point  of  Application  of  the  Resultant. 

155.  Catenary — Examples. 

§  IV.  the  pulley •-■•  88 

156.  Definition  of  the  Pulley— Fixed  and  Movable. 

157.  Use  of  Fixed  Pulley. 

158.  Equilibrium  in  single  Movable  Pulley. 

159.  Systems  of  Pulleys. 

160.  Equilibrium  in  first  System. 
161    Equilibrium  in  the  second. 

If2.  Equilibrium  in  the  third — Examples. 


CONTENTS.  IX 

Art  P"gl 

§   V.   THE    INCLINED  PLANE 108 

163.  Definition  of  Inclined  Plane. 

164.  Equilibrium  when  the  Body  is  sustained  by  a  Force  acting  in  any  Direction. 

165.  Equilibrium  when  the  Body  is  sustained  by  a  Force  parallel  to  the  Plane. 

166.  Equilibrium  when  the  Body  is  sustained  by  a  Force  parallel  to  the  Base  of  the  Plane. 

167.  Equilibrium  when  two  Bodies  rest  on  two  Inclined  Planes. 

§  VI.   THE  WEDGE 103 

168.  Definition  of  the  Wedge — Faces,  Angle   Back. 

169.  Conditions  of  Equilibrium  in  the  Wedge. 

170.  Defect  in  the  Theory. 

171.  Illustrative  Problem. 

§  VII.  THE  screw 103 

172.  Definition  of  the  Screw. 

173.  Conditions  of  Equilibrium  in  the  Screw. 

$  VIII.   BALANCES  AND   COMBINATIONS  OF  THE  MECHANICAL   POWERS.. 107 

174.  The  common  Balance. 

175.  Requisites  for  good  Balance. 

176.  Conditions  of  Horizontality  of  the  Beam. 

177.  Conditions  of  Sensibility. 

178.  Conditions  of  Stability. 

179.  Relations  of  the  Requisites. 

180.  Steelyard  Balance. 

181.  Law  of  Graduation. 

182.  Bent  Lever  Balance. 

183.  Law  of  Graduation. 

184.  Roberval's  Balance. 

185.  Condition  of  Equilibrium. 

186.  Conditions  of  Equilibrium  in  a  Combination  of  Levers. 

187.  Conditions  of  Equilibrium  in  the  Endless  Screw. 

188.  Conditions  of  Equilibrium  in  any  Combination  of  the  Mechanical  Powers- 

189.  Conditions  of  Equilibrium  in  the  Knee. 

*—  CHAPTER  VII. 

APPLICATION   OF   THE   PRINCIPLE   OF  VIRTUAL  VELOCITIES   TO   THE   MECHANICAL 
POWERS 116 

190.  Preliminary  Considerations. 

191.  Application  to  the  Wheel  and  Axle. 

192.  Application  to  Toothed  Wheels. 

193.  Application  to  Movable  Pulley  with  Parallel  Cords. 

194.  Application  to  the  first  System  of  Pulleys. 

195.  Application  to  the  second  System. 

196.  Application  to  the  third  System. 

197.  Application  to  the  Inclined  Plane. 

198.  Application  to  the  Wedge. 

199.  Application  to  the  Lever  of  any  Form. 

200.  Application  to  the  single  Movable  Pulley  with  inclined  Cords. 

CHAPTER  VIII. 

FRICTION 134 

201.  Definition  of  Friction— Kinds. 

202.  Measurement  of  Friction. 

203.  Laws  of  Friction. 

S04.  Value  of  the  Coefficient  of  Friction. 


*  CONTENTS. 

Art.  Pif. 

205.  Limits  of  the  Ratio  of  the  Power  to  the  Weight  ou  the  Inclined  Plane. 

206.  Limits  of  the  Ratio  of  the  Power  to  the  Weight  in  the  Screw. 

207.  One  Limit  obtained  directly  from  the  other. — Examples. 
203.  Examples  on  Chapters  VI.,  VII.,  and  VIII. 


DYNAMICS. 

INTRODUCTION *..  181 

209.  In  Dynamics,  Time  an  Element. 

210.  Definition  of  Motion. 

211.  Definition  of  Absolute  Motion. 

212.  Definition  of  Relative  Motion. 

213.  Definition  of  Velocity — Its  Measure. 

214.  Definition  of  Variable  Velocity — Its  Measure. 

215.  Definition  of  Relative  Velocity. 

216.  Definition  of  Inertia — First  Law  of  Motion. 

217.  Definition  of  Center  of  Inertia. 

218.  Definition  of  the  Path  of  a  Body. 

219.  Definition  of  Free  and  Constrained  Motion. 

220.  Definition  of  an  Impulsive  Force. 

221.  Definition  of  an  Incessant  Force. 

222.  Definition  of  a  Constant  Force — Its  Measure. 

223.  Definition  of  a  Variable  Force — Its  Measure. 

224.  Definition  of  Momentum — Its  Measure — Of  Living  Force. 

225.  Definition  of  a  Moving  Force — Its  Measure. 

226.  The  second  Law  of  Motion. 

227.  The  third  Law  of  Motion. 

CHAPTER  I. 

UNIFORM    MOTION j 3fi 

228.  Point  to  which  the  Force  must  be  applied. 

229.  General  Equation  of  Uniform  Motion. 

230.  Relation  of  Spaces  to  Velocities  when  the  Times  are  Equal. 

231.  Relation  of  Spaces  to  the  Times  when  the  Velocities  are  Equal. 

232.  Relation  of  Velocities  to  the  Times  when  the  Spaces  are  EquaL 

233.  Measure  of  an  Impulsive  Force. 

234.  The  Velocity  resulting  from  the  Action  of  several  Forces. 

235.  Parallelogram  of  Velocities. 

236.  Rectangular  Composition  and  Resolution  of  Velocities. 

237.  Relations  of  Space,  Time,  and  Velocity  of  two  Bodies  moving  in  the  same  Straight 

Line. 

238.  Relations  of  Space,  Time,  and  Velocity  of  two  Bodies  moving  in  the  Circumference 

of  a  Circle. 

239.  Examples. 

CHAPTER  II. 

IMPACT     OF    BODIES 143 

210.  Definition  of  Direct,  Central,  and  Oblique  Impact. 

241.  Definition  of  Elasticity — Perfect — Imperfect — Its  Modulus. 

242.  Definition  of  Hard  and  Soft. 

243.  Velocity  of  two  Inelastic  Bodies  after  Impact. 

244.  Loss  of  Living  Force  in  the  Impact  of  Inelastic  Bodies. 

245.  Velocities  of  imperfectly  Elastic  Bodies  after  Impact. 


CONTENTS.  X 

Art.  Pa^t 

246.  Velocity  of  the  nth  Body  in  a  Series  of  perfectly  Elastic  Bodies. 

247.  Velocity  of  the  common  Center  of  Gravity  before  and  after  Impact. 

248.  Conservation  of  the  Motion  of  the  Center  of  Gravity. 

249.  Definition  of  Angles  of  Incidence  and  Reflection. 

250.  Motion  of  an  Inelastic  Body  after  Oblique  Impact  on  a  Hard  Plane. 

251.  Motion  of  an  Elastic  Body  after  Oblique  Impact  on  a  Hard  Plane. 

252.  Direction  of  Motion  before  Impact,  that  a  Body  after  Impact  may  pass  through  a 

given  Point. 

253.  Measure  of  the  Modulus  of  Elasticity. 

254.  Mode  of  determining  it — Table  of  Moduli. 

255.  Examples. 

CHAPTER  III. 

MOTION  FROM   THE  ACTION   OF  A  CONSTANT    FORCE 153 

256.  Uniformly  accelerated  Motion — Acquired  Velocity. 

257.  Space  in  Terms  of  the  Force  and  Time. 

258.  Space  in  Terms  of  the  Force  and  Velocity. 

259.  Space  described  in  the  last  n  Seconds. 

260.  The  Velocity  and  Space  from  the  joint  Action  of  a  Projectile  and  Constant  Force. 

261.  The  Velocity  when  the  Space  is  given. 

262.  Velocity  lost  and  gained  by  the  Action  of  a  Constant  Force  when  the  Space  is 

the  same. 
{(63.  Scholium  on  Universal  Gravitation. 

264.  Scholium  on  the  Numerical  Value  of  the  Force  of  Gravity. 
°.65.  Examples. 

'  CHAPTER  IV. 

P  ROJ  E  CT  I  L  E  3 161 

266.  The  Path  of  a  Projectile  is  a  Parabola. 

267.  Equation  of  the  Path  when  referred  to  Horizontal  and  Vertical  Axes. 

268.  Definition  of  Horizontal  Range — Time  of  Flight — Impetus. 

269.  Time  of  Flight  on  a  Horizontal  Plane. 

270.  Range  on  a  Horizontal  Plane — The  same  for  two  Angles  of  Elevation. 

271.  Greatest  Height. 

272.  Range  and  Time  of  Flight  on  an  Inclined  Plane,  and  Co-ordinates  of  Point  of  Impact 

273.  Formula  for  Velocity  of  a  Ball  or  Shell. 

274.  Examples. 

CHAPTER  V. 

CONSTRAINED    MOTION. 
$  I.   MOTION   ON   INCLINED   PLANES 169 

275.  Relations  of  Space,  Time,  and  Velocity. 

276.  Velocity  down  the  Plane  and  its  Height. 

277.  Times  down  Inclined  Planes  of  the  same  Height. 

278.  Relations  of  Space,  Time,  and  Velocity  when  projected  up  or  down  the  Plane 

279.  Time  of  Descent  down  the  Chords  of  a  Circle. 

280.  Straight  Line  of  quickest  Descent  from  a  Point  within  a  Circle  to  its  Circumference. 

281.  Straight  Line  of  quickest  Descent  from  a  given  Point  to  an  Inclined  Plane. 

282.  Motion  of  two  Bodies  suspended  by  a  Cord  over  a  Fixed  Pulley. 

283.  Motion  of  two  Bodies  when  the  Inertia  of  the  Pulley  is  considered. 

§  II.   MOTION   IN   CIRCULAR  ARCS 173 

284.  Velocity  acquired  down  the  Arc  of  a  Circle. 

285.  Velocity  lost  in  passing  from  one  Side  of  a  Polygon  to  the  next 


\ 


XU  CONTENTS. 

Art.  Page 

286.  Velocity  lost  when  the  Sides  are  Infinite  in  Numler. 

267.  Direction  and  Intensity  of  an  Impulse  at  each  Angle,  to  make  a  Body  describe  a 
Polygon  with  a  uniform  Velocity. 

288.  Direction  and  Intensity  when  the  Polygon  becomes  a  Circle. 

289.  Definition  of  Centrifugal  and  Centripetal  Force. 

290.  Discussion  of  the  Motion  of  a  Body  in  a  Circle  by  the  Action  of  a  Central  Force. 

291.  Centrifugal  Force  compared  with  Gravity. 

292.  Centrifugal  Force  of  the  Earth  at  the  Equator. 

293.  Ratio  of  Centrifugal  Force  of  the  Earth  to  Gravity. 

294.  Time  of  its  Rotation  when  they  are  Equal. 

295.  Diminution  of  Gravity  in  different  Latitudes  by  the  Centrifugal  Force. 

296.  Cause  and  Value  of  the  Compression  of  the  Earth. 

297.  Centrifugal  Force  of  the  Moon  in  its  Orbit. 

298.  Moon  retained  in  its  Orbit  by  Gravitation. 

§  III.  pendulum 183 

299.  Definition  of  a  Simple  Pendulum. 

300.  The  Force  by  which  it  is  urged  in  its  Path. 

301.  Times  of  Descent  to  the  Center  of  Force  equal  when  the  Force  varies  as  the  Dia- 

tance. 

302.  Expression  for  the  Time  given. 

303.  Discussion  of  the  Vibrations  of  Pendulums. 

304.  Pendulum  used  to  determine  the  Figure  of  the  Earth. 

305.  Lengths  of  Pendulums  as  the  Squares  of  the  Number  of  Vibrations  in  a  given  Time. 

306.  Mode  of  determining  the  Length  of  the  Seconds  Pendulum. 

307.  Determination  of  the  Intensity  of  Gravity. 

308.  Correction  of  the  Length  of  the  Pendulum  for  a  given  Loss  or  Gain. 

309.  Rate  at  a  given  Height  above  the  Earth's  Surface. 

310.  Heights  determined  by  the  Change  of  Rate. 

311.  Definition  of  Conical  Pendulum. 

312.  Tension,  Velocity,  and  Time  determined. 

313.  Formula  for  Length  of  Pendulum  in  a  given  Latitude. 

314.  Examples. 

CHAPTER  VI. 

ROTATION   OF    RIGID   BODIES 19i 

315.  Angular  Velocity  of  a  Body  about  a  Fixed  Axis.  ' 

316.  Definition  of  Moment  of  Inertia. 

317.  Moment  of  Inertia  about  any  Axis  compared   with  that  about  a  Parallel  Axis 

through  the  Center  of  Gravity. 
,  318.  Radius  of  Gyration — Principal  Radius. 

319.  Existence  of  a  Point  at  which,  if  all  the  Matter  of  the  Body  were  collected,  the 

Motion  would  be  the  same. 

320.  Expression  for  its  Distance  from  the  Axis. 

321.  Definition  of  Center  of  Oscillation. 

322.  Time  of  Oscillation  of  a  Rigid  Body. 

323.  Definition  of  Compound  Pendulum. 

324.  Centers  of  Oscillation  and  Suspension  convertible. 

325.  Length  of  an  Equivalent  Simple  Pendulum. 

326.  Value  of  Principal  Radius  of  Gyration. 

327.  Relations  of  the  Centers  of  Gravity,  of  Gyration,  and  of  Oscillation. 

328.  Determination  of  the  Length  of  the  Seconds  Pendulum. 

329.  Examples  on  the  Relation  of  the  Simple  to  the  Compound  Pendulum. 

330.  Examples  on  the  Determination  of  the  Moment  of  Inertia. 


CONTENTS.  Xll] 

Art  Pag« 

CHAPTER  VII 201 

331.  More  General  Methods  required. 

332.  Modification  and  Extension  of  Fundamental  Formulae. 

§  I.    RECTILINEAR  MOTION  OF  A  FREE  POINT 202 

333.  Preliminary  Considerations — Absolute  Force. 

334.  Spac«  in  Terms  of  the  Time  when  the  Force  is  Constant. 

335.  General  Expression  for  the  Velocity  when  the  Force  is  Variable. 

336.  General  Expression  for  the  Time  when  the  Force  is  Variable. 

337.  Velocity  when  the  Force  varies  as  the  Distance. 

338.  Velocity  at  the  Center  of  the  Earth. 

339.  Time  when  the  Force  varies  as  the  Distance. 

340.  Velocity  when  the  Force  varies  as  the  Square  of  the  Distance. 

341.  Velocity  from  an  Infinite  Distance. 

342.  Time  when  the  Force  varies  inversely  as  the  Square  of  the  Distance. 

343.  Velocity  and  Time  when  the  Force  varies  inversely  as  the  Cube  of  tie  I»ista«ce. 

§  II.    CURVILINEAR  MOTION  OF  A  FREE  POINT 2i« 

344.  Preliminary  Considerations. 

345.  Motion  of  a  Point  acted  on  by  any  Number  of  Forces  in  £.e  name  Plane. 

346.  Specific  Directions  for  particular  Cases. 

347.  Velocity  of  a  Point  in  its  Path. 

348.  The  Velocity  may  be  found  when  the  Expression  id  i-ategrable. 

349.  The  Expression  integrable  when  the  Forces  are  directed  to  Fixed  Centers,  and 

are  Functions  of  the  Distances  from  those  Centers. 

350.  Motion  of  a  Point  when  the  Force  is  directed  to  a  Center. 

351.  When  there  is  an  Equable  Description  of  Areas,  the  Force  is  directed  to  a  Center. 

352.  Motion  of  a  Point  from  an  Impulsive  Force. 

353.  Motion  of  a  Projectile  acted  upon  hy  Gravity. 

§111.   CONSTRAINED   MOTION   O?   A   POINT 215. 

354.  Velocity  of  a  Point  on  a  giveu  Curve. 

355.  Time  of  its  Motion. 

356.  The  Reaction  of  the  Curve- 

357.  Motion  down  the  Arc  of  a  Circle  by  Gravity. 

358.  Motion  down  the  Are  of  a  Cycloid. 

359.  Tautochronism  of  Vibrations  in  a  Cycloid. 

§  IV.   MOTION    Of   A  POINT   ACTED   ON   BY  A   CENTRAL    FORCE 221 

360.  Preliminary  Remarks. 

361.  Expressions  fof  the  Force  in  Terms  of  Polar  Co-ordinates. 

362.  Equable  Description  of  Areas. 

363.  Angular  Velocity  of  the  Radius  Vector. 

364.  Velocity  of  a  Point  in  its  Orbit. 

365.  Time  of  describing  any  Portion  of  its  Orbit. 

366.  Equation  of  the  Orbit. 

367.  Geometrical  Values  of  Differential  Expressions. 

368.  Kepler's  Laws. 

369.  Accelerating  Force  of  a  Planet  directed  to  the  Sun. 

370.  Accelerating  Force  of  a  Planet  varies  inversely  as  the  Square  of  the  Distance. 

371.  Absolute  Force  the  same  for  all  the  Planets. 

372.  Velocity  of  a  Planet  at  any  Point  of  its  Orbit. 

373.  Determination  of  the  Orbit  when  the  Force  varies  inversely  as  the  S(pnare  of  th© 

Distance. 

374.  Time  of  describing  any  Portion  of  the  Orbit. 


<1V  CONTENTS. 

Art.  Pagl 

375.  Value  of  the  Absolute  Force  in  Terms  of  the  Masses—  In  Terms  of  Terrestria. 

Gravity. 
373,  Examples. 


HYDROSTATICS. 

CHAPTER  1 836 

377.  Forms  of  Matter. 

378.  Definition  of  a  Perfect  Fluid. 

379.  Definition  of  an  Incompressible  Fluid. 

380.  Definition  of  a  Compressible  Fluid. 

381.  Unimpaired  Transmission  of  Pressures. 

382.  Surface  of  a  Fluid  at  Rest. 

383.  Surface  of  a  Fluid  moving  Horizontally. 

384.  Surface  of  a  Fluid  Revolving. 

385.  Pressure  on  the  Horizontal  Base  of  a  Vessel. 

386.  Pressure  on  any  Side  of  a  Vessel  or  immersed  Surface. 

387.  Pressure  in  a  definite  Direction. 

388.  Resultant  Pressure  on  the  Interior  Surface  of  a  Vessel. 

389.  Resultant  Pressure  on  an  immersed  Solid. 

390.  Conditions  of  Equilibrium  of  an  immersed  Solid. 

391.  Definition  of  Plane  of  Flotation — Of  Axis  of  Flotation. 

392.  Depth  of  Flotation  of  a  Body. 

393.  Position  of  Equilibrium  of  a  Triangular  Prism. 

394.  Definition  of  Center  of  Pressure. 

395.  Formula  for  finding  the  Center  of  Pressure. 

396.  Center  of  Pressure  of  a  Rectangle  with  one  Side  in  the  Surface  of  the  Fluid. 

397.  Center  of  Pressure  of  a  Triangle  with  the  Vertex  in  the  Surface. 

398.  Center  of  Pressure  of  a  Triangle  with  a  Side  in  the  Surface. 

399.  Center  of  Pressure  of  a  Rectangle  wholly  immersed. 

400.  Equilibrium  of  Fluids  of  different  Density  in  the  same  Vessel. 

401.  Equilibrium  of  Fluids  of  different  Density  in  a  bent  Tube. 

402.  Examples. 

CHAPTER  II. 

SPECIFIC    GRAVITY -    ...    36« 

403.  Definition  of  Specific  Gravity. 

404.  Specific  Gravity  of  a  Body  more  Dense  than  Water. 

405.  Specific  Gravity  of  a  Body  less  Dense  than  Water. 

406.  Specific  Gravity  of  a  Liquid. 

407.  Weights  of  the  Constituents  of  a  Mechanical  Mixture. 

408.  Hydrometer.    1°.  Of  Constant  Weight.    2°.  Of  Constant  Volume. 

409.  Nicholson's  Hydrometer. 

410.  Examples. 

t 

CHAPTER  III. 

COMPRESSIBLE  OR  AERIFORM   FLUIDS 279 

■ill.  Tension  of  a  Compressible  Fluid  found. 

412.  The  Unit  of  Pressure. 

413.  Tension  inversely  as  the  Volume. 

414.  Tension  directly  as  the  Density. 


CONTENTS.  XX 

irt.  Pag, 

415.  Effect  of  Heat  on  Volume  and  Tension. 

415.  Relation  of  Density  to  Temperature  and  Pressure. 

416.  Density  of  the  Atmosphere  in  Terms  of  the  Height. 

417.  Barometrical  Measurement  of  Heights. 

418.  Examples. 


HYDRODYNAMICS. 

419.  Velocity  of  a  Fluid  in  a  Tube  of  variable  Diameter 22) 

420.  Velocity  of  a  Fluid  from  a  small  Orifice  in  the  Bottom  of  a  Vessel. 

421.  Horizontal  Range  of  a  Spouting  Fluid. 

422.  Quantity  of  Discharge  from  a  small  Orifice. 

423.  Time  required  for  a  Vessel  to  empty  itself. 

424.  Vena  Contracta — Coefficient  of  Efflux. 

425.  Discharge  from  a  Rectangular  Aperture. 

426.  Discharge  from  a  Triangular  Aperture. 

427.  Velocity  of  Efflux  of  an  Elastic  Fluid. 

428.  Motion  of  Fluids  in  Long  Pipes. 

429.  General  Method  of  determining  the  Discharge  from  small  Orifices. 

430.  General  Method  of  determining  the  Time  for  a  Vessel  to  empty  itself. 

431.  General  Method  of  determining  the  Discharge  from  Orifices  of  any  Form. 

432.  Examples. 


HYDROSTATIC  AND  HYDRAULIC  INSTRUMENTS. 

433.  Mariotte's  Flask 885 

434.  Bramah's  Press. 

435.  Hydrostatic  Bellows. 

436.  Diving  Bell. 

437.  Sea  Gage. 

438.  Siphon. 

439.  Common  Pump. 

440.  Common  Pump,  Conditions  of  Failure. 

441.  Air  Pump. 

442.  Condenser. 

443.  Clepsydra. 


ELEMENTARY   MECHANICS, 


INTRODUCTION. 

1.  Mechanics  is  the  Science  which  treats  of  the  laws  of 
Equilibrium  and  Motion.  It  is  subdivided  into  Statics,  Dy- 
namics, Hydrostatics,  and  Hydrodynamics. 

Statics  treats  of  the  necessary  relations  in  the  intensities  and 
directions  of  forces,  in  order  to  produce  equilibrium  of  solid 
bodies. 

Dynamics  treats  of  the  effects  of  forces  on  solid  bodies  when 
motion  is  produced. 

Hydrostatics  investigates  the  conditions  of  equilibrium  in 
fluid  bodies. 

Hydrodynamics  investigates  the  effects  of  forces  on  fluids 
when  motion  results.* 

2.  Force  is  that  which  produces  or  tends  to  produce  motion 
or  change  of  motion. 

The  consideration  of  the  nature  of  force  does  not  belong  to 
the  present  subject.  Mechanics  is  concerned  only  with  the 
effects  of  force  as  exhibited  in  the  production  of  motion  or  rest- 

3.  The  effect  of  a  force  depends  on,  1st,  its  Intensity ;  2d,  its 
Direction  ;  and,  3d,  its  Point  of  Application. 

The  Intensity  of  a  force  may  be  measured,  statically,  by  the 
pressure  it  will  produce,  or  by  the  weight  which  will  counter- 
poise it ;  dynamically,  by  the  quantity  of  motion  it  will  produce. 
By  assuming  for  a  unit  of  force  that  force  which  is  counter 
poised  by  a  known  weight,  the  intensity  or  magnitude  of  any 
other  force  will  be  expressed  by  the  numerical  ratio  which  its 
counterpoise  will  bear  to  the  counterpoise  of  the  unit  of  force. 

*  Theoretic  Statics  and  Dynamics  arc  those  branches  of  theoretic  Mechanics  which 
treat  of  the  effects  of  forces  applied  to  material  points  or  particles  regarded  as  without 
weight  or  magnitude.  Static  and  Dynamic  Somatology  would  then  embrace  the  appli- 
cation of  theoretic  Statics  and  Dynamics  to  bodies  of  definite  form  and  magnitude,  both 
(solid  and  fluid. 

A 


2  INTRODUCTION. 

In  the  saute  manner,  by  fixing  on  any  line  to  represent  the  unit 
of  force,  any  other  force  will  be  represented  bv  the  line  which 
bears  to  the  linear  unit  the  same  ratio  which  the  force  in 
question  bears  to  the  unit  of  force. 

The  Direction  of  a  force  is  the  line  which  a  material  point, 
acted  on  by  that  force,  would  describe  were  it  perfectly  free. 

The  point  of  application  of  a  force  is  that  point  in  its  line  ol 
direction  on  which  the  force  acts. 

4.  As  the  magnitudes,  directions,  and  points  of  lines  are  all 
determinable  by  the  principles  of  Analytical  Geometry,  so 
forces,  of  which  lines  are  the  appropriate  representatives, 
come  under  the  dominion  of  the  same  principles.  The  appli- 
cation of  these  principles  to  the  determination  of  the  laws  of 
equilibrium  and  motion,  considered  as  the  effects  of  forces, 
constitutes  analytical  mechanics. 

5.  Forces  may  act  on  a  point  either  by  pushing  or  pulling  it. 
As  these  are  readily  convertible,  the  one  into  the  other,  with- 
out affecting  the  intensity,  direction,  or  point  of  application  of 
the  forces  themselves,  all  forces  will  be  regarded  as  pulling 
unless  otherwise  expressly  stated. 

For  convenience,  we  shall  call  those  forces  which  have  a 
common  point  of  application  concurring  forces,  and  those 
which  act  along  the  same  line  toward  the  same  parts,  con- 
spiring forces. 

6.  A  body  is  an  assemblage  of  material  points.  The  material 
points,  or  elementary  particles,  are  connected  together  in  vari- 
ous ways,  according  to  the  nature  of  the  body. 

A  body  is  said  to  be  rigid  when  the  relative  position  of  its 
particles  remains  unchanged  by  the  action  of  forces  upon  it. 

In  flexible  and  elastic  bodies  the  relative  positions  of  the 
particles  change  by  the  action  of  forces. 


STATICS. 

CHAPTER    I. 

ON    THE    COMPOSITION    AND    EQUILIBRIUM    OF    CONCURRING    FORCES. 

7.  Proi\   Two  equal  forces  applied  to  the  same  point  in  ex- 
actly opposite  directions  are  in  equilibrium. 

For  no  reason  can  be  assigned  why  motion  should  take 
place  in  the  direction  of  one  force  which  will  not  equally  ap 
ply  to  the  other. 

8.  Prot.   Two  concurring  forces  forming  an  angle  with  each 
other  can  not  be  in  equilibrium. 

If  possible,  let  the  two  forces  P  and  Q,  acting  at  A,  and 
making,  by  their  directions,  the  an-     p" 
gle  PAQ,  be  in  equilibrium.     Apply 

to  the  point  A  the  force  P,,  equal  and  -< ^ > 

opposite  to  Q.     Since,  by  hypothe-   Pl  Q 

sis,  the  forces  P  and  Q,  are  in  equilibrium,  the  point  A  will 
move  in  the  direction  of  P,.  But  P,  and  Q,  being  equal  and 
opposite  {Art.  7),  will  be  in  equilibrium,  and  hence  the  point 
A  must  move  in  the  direction  of  P.  Therefore  the  point  A 
must  move  in  two  directions  at  once,  which  is  absurd. 
-  9.  Definition.  The  resultant  of  two  or  more  forces  is  a 
force  which  singly  will  produce  the  same  mechanical  effect  as 
the  forces  themselves  jointly. 

The  original  forces  are  called  components. 

Cor.  In  all  statical  investigations  the  components  may  be 
replaced  by  their  resultant,  and  vice  versa. 

10.  Prop.   The  resultant  of  several  conspiring  forces  is  a 
single  force  equal  to  their  sum.  and  acting  in  the  same  direction. 


STATICS. 


This  is  obvious  from  the  admitted  mode  of  measuring  forces. 

11.  Prop.  The  resultant  of  two  unequal  forces  acting  in  op~ 
posite  directions  is  a  single  force  equal  to  their  difference,  and 
acting  in  the  direction  of  the  larger. 

For  the  smaller  will  obviously  be  expended  in  annulling  in 
the  larger  a  quantity  equal  to  itself,  and  thus  leave  an  effective 
balance  equal  to  the  excess  of  the  larger  over  the  smaller,  and 
•icting  in  the  direction  of  the  larger. 

12.  Prop.  The  resultant  of  any  number  of  forces  acting  in 
the  same  right  line  is  equal  to  their  algebraic  sum. 

By  Art.  10,  the  resultant  of  each  system  of  conspiring  for- 
ces is  equal  to  their  sum,  and  by  Art.  11,  the  resultant  of 
these  two  resultants  is  equal  to  their  difference  with  the  sign 
of  the  greater  prefixed. 

13.  Prop.  In  a  system  of  points  invariably  connected,  any 
point  in  the  direction  of  a  force  may  be  taken  as  the  point  of 
application. 

If  the  force  P  be  applied  at  the  point  B,  and  to  any  othei 
.  r>  p    point  A  in  its  direction  two 

j j — >■     other    forces    be    applied, 

each  equal  to  it,  but  opposite  to  each  other,  since  B,  by  hypoth- 
esis, is  invariably  connected  with  A,  one  of  these  forces  {Art. 
7)  will  be  in  equilibrium  with  P.  There  will  remain,  there- 
fore, a  single  effective  force  equal  to  P  and  applied  at  A.  In 
the  same  manner  it  may  be  transferred  to  any  other  point  in 
its  direction. 

14.  Prop.  The  resultant  of  several  concurring  forces  in  one 
plane  lies  in  the  same  plane. 

For  if  we  suppose  the  resultant  to  lie  out  of  the  plane  of  the 
forces  on  one  side,  we  may  always  conceive  a  line  symmet- 
rically situated  on  the  other  side  of  the  plane  ;  and  since  no 
reason  can  be  assigned  why  it  should  be  in  one  of  these  lines 
rather  than  in  the  other,  it  can  be  in  neither  of  them,  unless  we 
admit  the  absurd  consequence,  that  it  is  in  both  ;  in  other 
words,  that  a  system  of  forces  has  two  resultants. 

15.  Prop.   The  resultant  of  two  equal  concurring  forces  is  in 


CONCURRING     FORCES.  5 

the  direction  of  a  line  bisecting  the  angle  formed  by  the  camp  > 
nents. 

For  no  reason  can  be  assigned  why  it  should  tend  to  one 
side  rather  than  the  other  of  this  line. 

16.  Cor.  When  the  forces  are  unequal,  it  is  obvious  that  the 
direction  of  the  resultant  will  make  a  less  angle  with  the  larger 
force  than  with  the  smaller,  and  the  greater  the  disparity  in 
the  forces,  the  smaller  will  be  this  angle. 

17.  Prop.  If  all  the  forces  of  a  system,  while  their  directions 
are  preserved,  are  increased  or  diminished  in  any  ratio,  their 
resultant,  without  changing  its  direction,  will  be  varied  in  the 
same  ratio ;  and  if  the  components  were  previously  in  equilib- 
rium, they  will  remain  so  in  whatever  ratio  their  intensities  be 
varied. 

Let  P,,  P2,  P3 be  any  concurring  forces  whatever. 

Since  (Cor.,  Art.  9)  we  can  replace  them  by  their  resultant 
R,  if  we  double  or  treble  each  of  the  components,  this  will  only 
double  or  treble  the  resultant  R,  without  changing  its  direction, 
because,  in  so  doing,  we  only  add  to  the  system  one  or  more 
systems  precisely  equal  to  the  first. 

In  like  manner,  if  we  reduce  the  original  components  to  one 
half  or  one  third  of  their  former  intensities,  the  resultant  will 
still  preserve  its  direction,  but  will  become  one  half  or  one 
third  as  large  as  before  ;  since  to  double  or  treble  these  is  only 
to  double  or  treble  their  resultant,  and  thus  reproduce  the 
original  system. 

If  the  forces  are  in  equilibrium,  by  varying  the  magnitudes 
of  all  in  the  same  ratio,  we  only  add  to  or  suppress  from  the 
system  other  systems  already  in  equilibrium. 

18.  Prof.  Three  equal  concurring  forces,  inclined  at  angles 
of  120°,  will  be  in  equilibrium. 

Since  each  force  is  inclined  in  the  same  Pl 
angle  to  the  directions  of  the  other  two, 
any  reason  that  can  be  assigned  why  ei- 
ther one  will  prevail,  will  apply  with 
equal  force  to  show  that  each  of  the  oth- 
ers will  prevail. 


D  STATICS. 

19.  Prop.  Two  equal  concurring  forces,  inclined  at  an  angle 
of  120° ,  have  for  their  resultant  a  force  which  will  be  repre- 
sented in  magnitude  and  direction  by  the  diagonal  of  a  rhombus 
constructed  on  the  lines  representing  the  components. 

Let  AD  and  AB  represent  the  two  forces  P,  and  P„,  mak- 
ing the  angle  BAD=120°.  To  the 
point  A  apply  a  force  P3  =  P,  or  P2, 
and  making,  with  Pj  and  P2,  angles  of 
120°.  By  Art.  18,  these  three  forces 
will  be  in  equilibrium.  Now  since  Pj 
and  P2  equilibrate  P3,  their  resultant 
must  also.  Therefore  produce  FA  to 
C,  making  AC=AF;  and  since  AC=R 
equilibrates  P3,  it  will  be  the  resultant  of  Pj  and  P2.  Join 
DC  and  BC,  and  ABCD  will  be  a  rhombus,  of  which  AC  is 
the  diagonal.  For  since  CAD=60°,  and  AC=AD,  the  triangle 
CAD  is  isosceles  ;  and  since  each  angle  is  equal  to  60°,  it  is 
equilateral.  .*.  DC=AD=AB.  In  the  same  manner  it  may  be 
shown  that  CB=AD,  and  hence  the  figure  is  a  rhombus. 

Con.  Denoting  the  angle  BAC  by  a,  we  have  AE  =  AB 
cos.  a=P  cos.  a,  and  the  resultant  AC  =  R=2P  cos.  a. 

20.  Prop.  If  tltree  forces  are  in  equilibrium,  each  will  be  equal 
and  opposite  in  direction  to  the  resultant  of  the  other  two. 

For,  replacing  either  two  by  their  resultant,  we  shall  have 
two  forces  in  equilibrium.  If  they  be  not  opposite  in  direc- 
tion, by  Art.  8,  they  can  not  be  in  equilibrium  ;  and  if  they  be 
not  equal,  by  Art.  11,  motion  will  ensue. 

PARALLELOGRAM    OF    FORCES. 

21.  Prop.  If  two  forces  be  represented  in  magnitude  and  di- 
rection by  the  two  adjacent  sides  of  a  parallelogram,  the  diago 
nal  will  represent  their  resultant  in  magnitude  and  direction. 

First.  The  direction  of  the  diagonal  is  that  of  the  resultant 

1°.  When  the  forces  are  equal,  this  is  obvious  from  Art.  15. 

2°.  Let  us  assume,  for  the  present,  that  it  is  also  true  for  the 

two  systems  of  unequal  forces  P  and  Q,  and  P  and  R ;  then  it 

will  be  true  for  the  forces  P  and  Q.+R-     Let  P  and  Q,  acl 


CONCURRING     FORCES. 


at  A  in  the  directions  AD  and  AB  respectively,  and  be  repre- 
sented in  magnitude  by  these  lines,  a      Q,      ^B      R      ^C 
Suppose  the  force  R  to  act  at  B 
(Art.  13),  a  point  in  its  direction, 
and  to  be  represented  in  magnitude 
by  BC.     Complete  the   parallelo-        D  E  F 

gram  ADFC,  draw  BE  parallel  to  AD,  and  the  diagonals  AE, 
AF,  and  BF.  Then  P  at  A,  in  AD,  and  Q  at  A,  in  AB,  have; 
by  hypothesis,  their  resultant  in  the  direction  AE,  and  (Art. 
13)  may  be  supposed  to  act  at  E.  Replacing  this  resultant  by 
its  components  acting  in  their  original  directions,  we  have  a 
force  P  acting  at  E  in  the  direction  BE,  and  a  force  Q  acting 
at  E  in  the  direction  EF.  Transfer  P  to  B  and  Q  to  F, 
without  changing  their  directions.  Then  P  at  B,  in  BE,  and 
R  at  B,  in  BC,  will  also,  by  hypothesis,  have  their  resultant  in 
the  direction  BF,  which  may  be  supposed  to  act  at  F.  We 
now  have  all  the  forces  acting  at  F,  and  this  without  disturb- 
ing their  effect  upon  the  point  A,  supposed  to  be  invariably 
connected  with  F.  Hence,  if  the  assumption  be  correct,  F  is 
a  point  in  the  direction  of  the  resultant  of  the  forces  P  and  Q 
+R.  But  the  assumption  is  correct  when  Q,  and  R  are  each 
equal  to  P  (Art.  15).  Therefore,  the  proposition  for  the  direc- 
tion of  the  resultant  is  true  for  P  and  2P.  Again,  making  Q 
=  2P  and  R=P,  it  is  true  for  P  and  3P,  and  so  for  P  and  wP. 
Also,  putting  nP  for  P,  Q,=P,  and  R  =  P,  it  will  be  true  for  ?iP 
and  2P,  and  so  on  for  mP  and  nY  (m  and  n  being  positive  in- 
tegers), or  for  all  commensurable  forces. 

3°.  When  the  forces  are  incommensurable.     Let  AB 
AC   represent   the   forces.      Complete  A  v  B 

the  parallelogram,  and  draw  the  diag- 
onal AD.  AD  will  represent  the  di- 
rection of  the  resultant.  If  not,  let 
some  other  line,  as  AE,  be  its  direc- 
tion.     Divide   AB  into   a  number   of 

equal  parts  less  than  DE,  and  on  AC  C  D 

take  as  many  of  these  parts  as  possible.  Since  AC  and  AB 
are  incommensurable,  there  will  be  a  remainder  GC  less  than 
DE.     Draw  GF  parallel  to  AB,  and  join  AF.     AB  and  AG 


nd 


8  STATICS. 

representing  two  commensurable  forces,  AF  will  represent  thj 
direction  of  their  resultant.  But  {Art.  16)  the  resultant  of  AB 
and  AC  will  make  a  less  angle  with  AC  than  the  resultant  of 
AB  and  AG  does.  .'.  AE  can  not  be  the  direction  of  the  re- 
sultant of  AB  and  AC  ;  and  similarly,  it  can  be  shown  that  no 
other  direction  than  AD  can  be  that  of  the  resultant. 

Second.  The  diagonal  will  represent  the  magnitude  of  the 

resultant.  Let  P,  Q,  and  R 
be  three  forces  in  equilibrium, 
and  AD,  AB,  AF  represent 
their  magnitudes  respective- 
ly. Complete  the  parallelo- 
grams BD  and  DF,  and  draw  the  diagonals  AC  and  AE.  AC 
being  the  direction  of  the  resultant  of  P  and  Q,,  must,  by  Art. 
20,  be  in  the  same  straight  line  with  AF,  and  is  therefore  par- 
allel to  DE.  In  the  same  manner,  AE  being  the  direction  of 
the  resultant  of  P  and  R,  will  be  in  the  direction  of  BA  pro 
duced,  and  therefore  parallel  to  CD.  Hence  CAED  is  a  par- 
allelogram, and  CA  is  equal  to  DE,  which  is  equal  to  AF,  by 
construction.  But,  by  Art.  20,  AF  represents  the  magnitude 
of  the  resultant  of  P  and  Q.  Hence  CA,  the  diagonal  of  the 
parallelogram  constructed  on  the  lines  AD  and  AB,  represents 
the  resultant  of  P  and  Q,  in  magnitude  as  well  as  direction. 

TRIANGLE    OF    FORCES. 

22.  Prop.  If  three  concurring  forces  are  in  equilibrium,  and 
a  triangle  be  formed  by  lines  draum  in  their  directions,  the  sides 
of  the  triangle,  taken  in  order,  will  represent  the  forces.  Con- 
versely, if  the  forces  can  be  represented  by  the  sides  of  a  triangle, 
taken  in  order,  they  will  be  in  equilibrium. 

Let  the  forces  P,  Q,  and  R  be  in  equilibrium,  and  be  repre 
sented  by  AB,  AD,  and  AE  respectively. 
Produce  EA,  draw  BC  parallel  to  AD, 
and  complete  the  parallelogram.  AC 
will  represent  the  resultant  of  P  ai.d  Q, 
{Art.  21) ;  and  since  R  equilibrates  P 
D  and  Q,  it  must  be  equal  and  oppos;,*.e  to 

their  resu.tant.      .'.  EA=AC.      And  since  BC  is  equal    and 


CONCURRING     FORCES.  9 

also  parallel  to  AD,  that  is,  has  the  same  direction,  the  three 
sides  of  the  triangle  ABC  taken  in  order  represent  the  three 
forces  P,  Q,,  and  R. 

Conversely.  If  the  three  sides  AB,  BC,  and  CA  of  the  tri- 
angle ABC,  taken  in  order,  represent  the  direction  and  magni- 
tude of  the  three  forces  P,  Q,  and  R,  they  will  be  in  equilibri- 
um. Draw  from  A  a  line  parallel  to  BC,  and  from  C  a  line 
parallel  to  AB,  meeting  the  former  in  D.  The  resultant  of 
the  two  forces  P  and  Q,  represented  by  AB  and  BC,  or  AB 
and  AD,  is  equal  to,  and  in  the  direction  of  the  diagonal  AC ; 
that  is,  equal  and  opposite  in  direction  to  the  force  R.  Hence 
P,  Q,  and  R  are  in  equilibrium. 

23.  Cor.  Hence  a  given  force  may  be  resolved  into  two 
component  forces,  acting  in  given  directions.  Also  into  two 
others,  one  of  which  is  given  in  magnitude  and  direction. 

1°.  Let  AB  be  the  given  force,  and 
AC,  AD  the  given  directions  ;  that  is, 
making  known  angles  with  AB.  From 
B  draw  BF,  and  BE  parallel  to  AD  and 
AC.  Then  AF  and  AE  are  the  two 
components  {Art.  21)  acting  in  the  given 
directions  AC  and  AD.  **     ^D 

2°.  To  resolve  AB  into  two  others,  one  of  which  shall  be  in 
the  direction  AC,  and  be  equal  to  AF.  Draw  FB,  and  com- 
plete the  parallelogram.     AE  will  be  the  other  component. 

"When  the  directions  of  the  components  are  arbitrary,  then- 
valuation  will  be  most  easily  effected  by  assuming  these  di- 
rections at  right  angles.  If  the  angle  CAD  be  right,  then  AE 
=AB  cos.  BAE,  and  AF=AB  cos.  BAF. 

POLYGON    OF    FORCES. 

24.  Prop.  If  any  number  of  concurring  forces  be  represented 
in  magnitude  and  direction  by  the  sides  of  a  polygon,  taken  in 
order,  they  will  be  in  equilibrium. 

Let  the  sides  of  the  polygon  ABCDEA  represent  the  mag- 
nitudes and  have  the  directions  respectively  of  the  forces  P, 
P2,  P3,  P4,  and  Ps.     These  forces  will  be  in  equilibrium. 


10 


STATICS. 


Draw  the  diagonals  AC  and  AD.     By  Art.  22,  AC,  the  third 

side  of  the  triangle  ABC,  repre- 
sents a  force  equivalent  to  the 
two  forces  AB  and  BC.  AC  is 
therefore  the  resultant  of  Pj  and 
P2,  and  may  be  substituted  for 
them.  In  the  same  manner,  AD 
is  equivalent  to  the  forces  AC 
*D  and  CD,  or  to  P„  P2,  and  P3  ; 
and  AE  to  the  forces  AD  and 
DE,  or  to  the  forces  Pl5  P2,  P3, 
and  P4.  That  is,  AE  is  the  re- 
sultant of  the  four  forces  P1?  P2,  P3,  and  P4,  and  acts  in  the 
direction  AE.  But  P5  is  represented  in  magnitude  by  EA,  and 
acts  in  the  direction  EA.  Hence  Ps  is  equal  and  opposite  to 
the  resultant  of  the  other  four,  or  the  forces  are  in  equilibrium. 
Cor.  The  proposition  is  true  of  forces  which  do  not  lie  all  in 
one  plane.     For  the  proof  is  independent  of  this  supposition. 

25.  Scholium.  When  three  forces  are  in  equilibrium,  any 
three  lines,  taken  parallel  to  their  directions,  will  form  a  trian- 
gle, the  sides  of  which  respectively  will  represent  the  relative 
magnitudes  of  the  forces ;  but  when  there  are  four  or  more 
forces  this  will  not  hold,  since  the  relation  which  subsists  be- 
tween the  sides  and  angles  of  triangles  does  not  obtain  in  poly 
gons  of  more  than  three  sides. 

26.  Prop.  To  find,  graphically,  the  resultant  of  any  number 
of  forces,  acting  at  different  points  in  the  same  plane. 

Let  the  forces  Px,  P2,  P3  have  their  points  of  application  at 
A,  B,  and  C.  Producing  the  directions  of 
the  forces  P,  and  P2  till  they  meet  in  D, 
construct  the  parallelogram  AB  on  the 
lines  representing  them,  and  the  diagonal 
will  represent  their  resultant  Ra.  Pro- 
ducing Rj  until  it  meets  the  direction  of 
P3  in  F,  and  constructing  the  parallelo- 
gram on  the  lines  representing  Rl  and  P3, 
the  diagonal  will  represent  their  resultant 


CONCURRING     FORCES.  11 

R2,  or  the  resultant  of  P1?  P2,  and  P3.  By  the  same  process 
we  may  find  the  resultant  of  any  number  of  forces.  If  any 
two  are  parallel,  we  must  first  compound  one  of  these  with  a 
third,  whose  direction  is  inclined  to  it.  If  all  are  parallel,  tne 
case  will  belong  to  parallel  forces. 

PARALLELOPIPED    OF    FORCES. 

27.  Prop.  If  three  concurring  forces  lying  in  different  planes 
be  represented  in  magnitude  and  direction  by  the  three  edges  of 
a  parallelopiped,  then  the  diagonal  will  represent  their  resultant 
in  magnitude  and  direction ;  and  conversely,  if  the  diagonal 
represents  a  force,  it  is  equivalent  to  three  forces  represented  by 
the  edges  of  the  parallelopiped. 

Let  the  three  edges  AB,  AC,  AD  of  the  parallelopiped  rep- 
resent the  three  forces.  Then  AE,  B 
the  diagonal  of  the  face  ACED, 
represents  the  resultant  of  the 
forces  AD  and  AC.  Compound- 
ing this  with  the  third  force,  rep- 
resented by  AB,  we  have  AF,  the  c~ 
diagonal  of  the  parallelogram  AEFB,  for  the  resultant  of  AE 
and  AB,  or  of  the  forces  AC,  AD,  AB. 

Reciprocally,  the  force  AF  is  equivalent  to  the  components 
AB,  AE,  or  to  the  components  AB,  AC,  and  AD. 

28.  Prop.  If  three  forces  are  in  equilibrium,  they  are  propor- 
tional each  to  the  sine  of  the  angle  made  by  the  directions  of  the 
other  two. 

By  Art.  22  and  figure, 

P  :  Q  :  R  =  AB  :  BC   :  CA, 
and  by  trig.,  =sin.  BCA  :  sin.  CAB  :  sin.  ABC  ; 

=  sin.CAD  :  sin.  CAB  :  sin.  ABC  ; 
or  because  sin.  A=  sin.  (180°  — A), 

=sin.  DAE  :  sin.  EAB  :  sin.  BAD  ; 
=sin.  QR    :  sin.  Pit     :  sin.  PQ. 

29.  Prop.  If  P  and  Q  be  two  concurring  forces,  6  the  angle 
made  by  their  directions,  and  R  their  resultant,  then  R2=Pa  + 
Q2+2PQ  cos.  6. 


12 


STATICS. 


By  trig,  we  have,  from  jig.,  Art.  22, 

AC2=AB2+BC2-2AB.BC  cos.  ABC, 
and  cos.  ABC=  — cos.  BAD=  — cos.  6. 

R2=P2+Q2+2PQcos.0. 

30.  Def.  The  moment  of  a  force  about  any  point  is  the  prod- 
uct of  the  force  into  the  perpendicular  let  fall  from  that  point 
on  the  direction  of  the  force.  The  point  is  called  the  origin 
of  moments. 

The  moment  of  a  force  measures  the  tendency  of  the  force 
to  produce  rotatory  motion  about  a  fixed  point. 

31.  Prop.  The  moment  of  the  resultant  of  two  forces  equals 
the  algebraic  sum  of  the  moments  of  the  components. 

1°.  When  the  origin  of  moments  falls  without  the  angle 
made  by  the  forces. 

Let  AB,  AC  represent  the  two 
forces  P  and  Q,  and  AD  the  diago- 
nal of  the  parallelogram  constructed 
on  AB  and  AC,  their  resultant  R. 
H  Take  any  point,  O,  for  the  origin  ol 
moments  ;  join  OA  and  draw  AG  per 
pendicular  to  OA ;  draw  Ol,  0>.*.,  On 
respectively  perpendicular  to  AB,  AC, 
AD,  and  BE,  CF,  DG  perpendicular 
to  AG ;  also  CH  parallel  to  AG. 

The  triangles  O/A,  OmA,  OnA  are  respectively  similar  to 
the  triangles  AEB,  AFC,  AGD. 

Whence      AE:AB=OZ 


G 


OA,  or  AE= 
AF  :  AC  =Om  :  OA,  or  AF= 
AG  :  AD=0?i  :  OA,  or  AG= 


AB.Ol 


OA  ' 
AC.Om 
~OA~~' 
AD.Orc 

OA    ' 


But  the  triangles  ABE  and  CDH  oeing  equal,  AE  =  CH  =  FG. 
.-.  AE+AF=AG, 
or  AB.0l+AC.0m=AD.0n,  or  designating  Ol,  Om,  and  On  by 
p,  q,  and  r  respectively,  Y.p+Q.q=Tl.r. 

2°.  When  the  origin  of  moments  is  taken  within  the  angle. 


CONCURRING     FORCES. 


13 


In  this  case  the  moments  of  the  forces  tend  to  produce  rota- 
tion in  opposite  directions.  As- 
suming one  direction  as  positive, 
to  distinguish  them  we  must  regard 
the  other  as  negative.  Let  the 
positive  direction  be  that  of  the 
hands  of  a  watch  ;  then  Q.Om  will 
be  positive,  and  P.O/  negative. 

The  proof  is  the   same   in  this 
case,  except  that  AG=AF— AE, 

and  AC.Om-AB.OZ=AD.Orc, 

or  Q-q— P.p=R.r. 

32.  Cor.  1.  The  moment  of  the  resultant  of  any  number  of 
concurring  forces  in  the  same  plane,  is  equal  to  the  algebraic  sum 
of  the  moments  of  the  components. 

By  compounding  the  resultant  R  with  a  third  force,  we 
should  obtain  a  like  result.  In  the  same  manner,  the  proposi- 
tion may  be  extended  to  any  number  of  forces. 

33.  Cor.  2.  If  the  origin  of  moments  be  a  fixed  point,  and 
taken  in  the  direction  of  the  resultant,  On  will  become  zero, 
and 

P.O/=Q.Om,  or  ?.p=Q,.q ; 
that  is,  while  the  fixed  point  O,  by  its  resistance,  counteracts 
the  resultant  force,  the  forces  P  and  Q,  will  be  in  equilibrium 
about  that  point,  since  their  moments,  tending  to  cause  rota- 
tion in  opposite  directions,  are  equal. 

34.  Cor.  3.  If  several  forces  are  in  equilibrium,  the  resultant 
force  R  is  zero,  and  the  moment  of  the  resultant  R.r=0. 
Hence  the  moments  in  one  direction  balance  those  in  the  op- 
posite direction,  and  there  is  no  tendency  to  motion  either  ot 
translation  or  rotation. 


35.    EXAMPLES. 

1.  When  the  component  forces  are  P  and  Q,  and  the  angle 
made  by  their  directions  6 ;  what  is  the  magnitude  of  the  re- 
sultant R  when  0—0  and  6=tt  1 

Ans.  (P+Q)  and  (P-Q). 


14  STATICS. 

2.  Show  that  R  is  greatest  when  6—0,  least  when  0— n,  and 
intermediate  for  intermediate  values  of  0. 

3.  When  P=Q  and  0=60°,  find  R. 

Ans.  R=PV3. 

4.  When  P=Q  and  0=135°,  find  R. 

Ans.  R=Y\J2-  V~2. 

5.  When  the  three  concurring  forces  3m,  4m,  and  5m  are 
in  equilibrium,  find  the  angle  3m, 4m. 

Ans.  90°. 

6.  If  P=Q  and  0=120°,  find  R. 

Ans.  R=P. 

7.  If  P=G,  Q=ll,and  0=30°,  find  the  magnitude  of  R,  am 
of  the  angles  P,R  and  Q,R. 

Ans.  R=  16.47,  P?R=19°.30',  q7R=10°.30'. 

8.  Apply  the  proof  of  the  polygon  of  forces  to  the  case  ot 
five  equal  forces  represented  by  the  sides  of  a  regular  penta 
gon  taken  in  order. 

9.  A  cord  is  tied  round  a  pin  at  the  fixed  point  A,  and  its 
two  ends  are  drawn  in  different  directions  by  the  forces  P  and 
Q.     Find  the  angle  P,Q=0,  when  the  pressure  on  the  pin  is 

R-P+Q 

2PQ-3(P+Q2) 

Ans.  Cos.  0= -pQ . 

10.  A  cord,  whose  length  is  21,  is  tied  at  the  points  A  a'nd  B 
m  the  same  horizontal  line,  whose  distance  is  2a;  a  smooth 
ring  upon  the  cord  sustains  a  weight  w :  find  the  force  T  of 
tension  in  the  cord. 

Ans.  T=- 


V-4 

11.  Given  the  four  concurring  forces  Pj  =  l,  P2=2,  P3=3, 
P4=4,  and  the  angles  P^P3  =  90°,  P^P4=90°,  and  V^?2  = 
60°.     Find  the  magnitude  of  the  resultant  R,  and  its  direction 

pTX 

Ans.  R=6.889  and  P^k=102°.16'. 


W 


&  z   t  \    W  " , 


CHAPTER   II. 


PARALLEL     FORCES. 

Having  considered  the  subject  of  forces  which  have  a  com- 
mon point  of  application,  or  which  are  reducible  to  this  condi- 
tion, we  next  proceed  to  consider  forces  which  act  on  differ- 
ent points,  connected  together  in  some  invariable  manner,  as 
in  rigid  bodies,  and  whose  directions  are  parallel. 

3G.  Prop.  The  resultant  of  two  parallel  forces,  acting  at  the 
extremities  of  a  rigid  right  line,  is  parallel  to  the  components, 
equal  to  their  algebraic  su?n,  and  divides  the  right  line,  or  the  right 
line  produced,  into  parts  reciprocally  proportional  to  the  forces. 

1°.  When  the  forces  act  in  the  same  direction. 

Let  the  parallel  forces  P  and  P!  act  at  A  and  B.  At  these 
points  apply  the  equal  and 
opposite  forces  P'  and  P'  l  ; 
these  will  not  disturb  the 
system.  P  and  P'  at  A 
will  have  a  resultant  P", 
and  Pj  and  P',  at  B,  a  re- 
sultant P",.  The  direc- 
tions of  P"  and  P",  will 
meet  in  some  point  D,  at 
which  we  may  suppose 
them  to  act.  Replacing  P"  by  P  and  P',  and  P",  by  P,  and 
P'j,  we  now  have,  acting  at  D,  the  four  forces  P,  P',  P,, 
and  P'j,  of  which  P'  and  P',  are  equal  and  opposite,  while  P 
and  Pj  act  in  the  direction  DC,  and  have  a  resultant  R  =  P+ 
Pj  {Art.  10).  To  determine  the  point  C  in  the  line  AB,  where 
R  acts,  since  the  sides  of  the  triangle  ACD  have  the  directions 
respectively  of  the  forces  P,  P',  and  their  resultant  P"  (Art 
22  V  we  have, 


>P 


16 


STATICS. 


P  :  P'=DC  :  CA, 

and  similarly  from  triangle  BCD, 

P',  :Pj=BC:DC. 
Compounding  these  ratios,  we  have,  since  P'—P',, 

P  :  P,=BC  :  CA,  or  P.AC=P,.BC. 
2°.  When  the  forces  act  in  opposite  directions. 
Let  P  and  Px  (Pj>P)  be  two  parallel  forces,  acting  at  A 

and  B  in  opposite  directions, 
and  apply  to  these  points,  aa 
before,  the  equal  and  opposite 
forces  P'  and  P'j.  The  re- 
sultant P"  of  P  and  P'  will 
meet  the  resultant  P",  of  P, 
and  P'j  in  some  point  D, 
since  (Art.  16)  P",  makes  a 
less  angle  with  P,  than  P" 
does  with  P.  We  shall  then 
have,  as  before,  the  four  forces 
P,  P,,  P',  and  P'j,  acting  at 
D,  of  which  P'  and  P'j  are  equal  and  opposite,  while  P  and  Pj 
are  opposite  and  unequal,  and  have  a  resultant  R=P,  —  P,  act- 
ing at  any  point  in  its  direction,  as  C  in  AB  produced. 

To  determine  the  point  C,  we  have  (Art.  22),  from  the  tri- 
angles ACD  and  BCD,  as  in  Case  1°, 

P     :  P'  -DC  :  CA 
and  P'j  :Pj=BC  :  CD  ; 

compounding  P     :  Pj=BC  :  CA,  or  P.AC  =  Pj.CB. 

37.  Def.  When  AB  is  perpendicular  to  the  direction  of  the 
forces,  AC  and  BC  are  called  the  arms  of  the  forces,  and  the 
products  P.AC  and  P,.CB  are  the  moments  of  the  forces  about 
the  point  C. 

38.  Cor.  If  to  the  point  C  we  apply  a  force  R,,  equal  and 
opposite  to  R,  the  forces  P,  Pj,  and  Rj  will  be  in  equilibrium 
about  that  point.  For  the  resultant  R,  passing  through  C,  will 
be  counteracted  by  R , ,  so  there  can  be  no  motion  of  transla- 
tion.    And  if  we  draw  through  C  a  line  perpendicular  to  the 


PARALLEL     FORCES.  17 

direction  of  the  forces,  calling  the  parts  intercepted  by  P  and 
Pj  respectively  p  and  px  ;  from  the  similar  triangles,  thus 
formed,  we  should  have 

BC:CA=^,  :p. 
.:  PiP^p,  :p,  or  P.jp=P1p1. 

Hence  the  moments  of  P  and  Pa,  which  measure  the  tendency 
to  rotation  in  opposite  directions,  being  equal,  there  can  be  no 
motion  of  rotation. 

39.  Prop.  To  determine  the  point  of  application  of  the  re- 
sultant in  terms  of  the  components  and  distance  between  their 
points  of  application. 

1°.  When  the  forces  act  in  the  same  direction.  By  Art.  36 
and  figure,  we  have 

P.AC=P1.BC=P1.(AB-AC), 

or  AC=F?^-.AB; 

and  P.AC=P1.BC=P(AB-BC), 

or  BC=FJ^-.AB. 

2°.  When  the  forces  act  in  opposite  directions,  we  have 
P.AC^Pj.BC^P^AC-AB), 

or  AC=p^-p.AB; 

P 
and  similarly,  BC=p p.AB. 

40.  Cor.  1.  When  the  forces  act  in  opposite  directions,  the 
resultant  lies  without  the  components  and  on  the  side  of  the 
larger. 

41.  Cor.  2.  When  the  forces  are  equal  and  opposite,  we 
have 

R^-P^O, 
or  there  will  be  no  motion  of  translation. 

Also,  AC=p^p.AB=^-.AB=x. 

B 


*<\ 


18 


STATICS. 


Hence  an  equilibrium  can  not  be  produced  except  by  the  ap« 
plication  of  an  infinitely  small  force  at  a  point  whose  distance 
is  infinite  ;  that  is,  two  equal  and  parallel  forces,  acting  in  op- 
posite directions,  can  have  no  single  resultant. 

42.  Def.  Such  forces  are  called  a  statical  couple.  Their  ef- 
fect is,  tendency  to  rotatory  motion  only,  and  all  tendency  to 
rotatory  motion  can  be  referred  to  forces  forming  such  couples. 

43.  Prop.  To  find  the  resultant  of  any  number  of  parallel 
forces  acting  at  any  points  in  the  same  plane. 

P2 


1°.  Let  the  parallel  forces  Pt,  P2,  P3  .  .  .  P„  all  act  in  the 
same  direction.     Draw  any  two  rectangular  axes  OX,  OY, 
and  let 
x,  y  ,  be  the  co-ordinates  of  A,  the  point  of  application  of  F^ 


x2  y2 

xnyn 
also, 
x'  y' 

x"  y" 


B, 
D, 

N, 

C, 
E, 


Po 

p3 

p„ 

R. 


x   y 


R, 


R. 


Draw  Ac  and  Cb  parallel  to  OX.     The  triangles  ACc  and 
CB6  are  similar,  and  give 

AC  :  BC=Cc  :  B&. 
By  Art.  36,  R^P^P,, 


PARALLEL     FORCES. 


19 


and       P,  :  P2=BC  :  AC=B&  :  Cc,  or  P1.Cc=Pa.B6, 
P1.(CH-AF)=P0.(BG-CH), 
(P,  +P2)CH=P1  AF+P2.BG ; 

Compounding  R,  with  P3,  we  shall  find  the  second  result- 
ant, 

Ra=R1+P,  =  P1+Pa+P„ 
and  R2.EL=R1.CH+P3.DK, 

R2.EL=P1AF+P2.BG+P3DK; 

™  R2y,=Piy1+P2y2+P3y3- 

By  continuing  the  same  process,  we  should  find,  ultimately, 

R=P1+Pa+P3  + Pn  (a), 

Ry=Piyi+P2y2+P3y3+  ....  ?„yn  (c). 

By  drawing  lines  from  A,  C,  B,  E,  D;  &c,  parallel  to  OX 
we  may  find,  similarly, 

Ri=P1x,+P2a;2+P3a:3+,  &c +Vnxn      (b). 

*  2°.  When  some  of  the  forces  act  in  opposite  directions.    To 
these  the  negative  sign  must  be  prefixed. 

If  Rj,  the  resultant  of  y 
Pj  and  P2,  as  found  in  the 
previous  case,  be  com- 
pounded with  P3  acting 
at  D,  in  a  direction  oppo- 
site to  that  of  Pj  and  P2, 
by  joining  CD  and  pro- 
ducing it  {Art.  40),  in  the 
direction  of  the  greater 
force,  say  R,,  we  have 
R2,  the  second  resultant. 

R2=Rj— P3,  and  E  being  its  point  of  application,  R,.EC 
-P3.ED.     Drawing  Ce,  Ec/,  parallel  to  OX,  and  meeting  DK 
in  d  and  LE  produced  in  e,  the  triangles  ECe,  EDc?  are  simi- 
lar ;  and 

EC  :  DE=Ee  :  Drf. 
.-.  R1.Ee=P3.Drf; 
or  R1(CH-EL)=P3.(DK-EL), 


w 


STATICS. 


or  (R1-P3).EL=R1.CH-P3.DK5 

or  R2.EL=P1.AF+P2BG-P3.DK, 

and  so  on  for  any  other  forces  acting  in  directions  opposite  to 
P,  and  P2. 

3°.  When  the  point  of  application  of  either  force  lies  in  any 
other  angle  than  YOX,  to  its  co-ordinates  must  be  given  their 
appropriate  sign. 

The  ordinate  y2  of  B,  the  point  of  application  of  P2,  will  be 
negative.     Draw  Aa,  Cb  parallel  to  OX. 

Y 

Pi 


TB 

P1.CA=P2.CB. 
By  sim.  triangles,  P!.Ca=P2.B&; 

or  P1(AF-CH)=P2(CH+GB), 

or  (P1+P2).CH=P1.AF-P2.BG, 

or  R^P^-P^. 

. .  The  general  formulae  (a),  (b),  (c)  will  apply  to  all  cases, 
by  giving  the  proper  signs  to  the  forces  and  to  the  co-ordinates 
of  their  points  of  application. 

These  formulae  are  more  concisely  written  by  using  the 
Greek  letter  2  as  the  sign  of  summation,  P  to  represent  either 
force,  and  x,  y  its  co-ordinates  ;  thus, 

R.=s(P)  (O, 

R..r  =  2(P.x)  (&'), 

R.y=2(P.y)  (C). 

When,  therefore,  the  magnitudes  of  the  forces  and  Jie  co-or 


PARALLEL     FORCES.  21 

dinates  of  their  points  of  application  are  given,  (a)  will  give 
the  magnitude  of  the  resultant,  and  (b)  and  (c)  the  co-ordinates 
x,  y  of  its  point  of  application. 

44.  Def.  The  point  whose  co-ordinates  are  x,  y  is  called  the 
center  of  parallel  forces.  Its  position  depends  on  the  magni- 
tude of  the  forces  and  the  co-ordinates  of  their  points  of  appli- 
cation, but  is  independent  of  their  common  direction ;  for,  by 
turning  the  forces  around  their  respective  points  of  applica- 
tion, at  the  same  time  preserving  their  parallelism,  it  will  be 
seen  that  the  position  of  the  point  E  is  not  thereby  affected. 

45.  Prop.  The  moment  of  the  resultant  of  any  number  of  par- 
allel forces,  acting  in  the  same  plane,  is  equal  to  the  algebraic 
<*um  of  the  moments  of  the  components. 

Since  the  origin  of  the  co-ordinate  axes  and  their  direction 
are  arbitrary  {see  figure  of  Art.  43),  suppose  the  axis  OX  to  be 
drawn  perpendicular  to  the  direction  of  the  forces,  and  the 
forces  produced  to  intersect  it.  Then  Pj^,  P2a:3,  &c,  will 
represent  the  moments  of  P,,  P2  respectively,  and  Rx  the  mo- 
ment of  the  resultant. 

46.  Def.  The  moment  of  a  force  with  reference  to  a  plane  is 
the  product  of  the  intensity  of  the  force  by  the  perpendicular 
let  fall  from  the  point  of  application  upon  the  plane.  Thus 
P,^,  is  the  moment  of  the  force  P,,  in  reference  to  the  plane 
passing  through  OY  perpendicular  to  OX,  and  P1yl  is  the 
moment  of  the  force  P,,  with  reference  to  the  plane  through 
OX  perpendicular  to  OY. 

47.  Prop.  To  determine  the  conditions  of  equilibrium  of  any 
number  of  parallel  forces. 

Suppose  the  forces  in  the  figure  {Art.  43)  all  turned  round 
their  points  of  application  so  as  to  become  perpendicular  to  the 
plane  of  YOX.  They  will  no  longer  be  in  the  same  plane,  but 
as  their  intensity  is  not  thus  changed,  nor  their  points  of  appli- 
cation, {a'),  (&'),  (c')  will  still  hold.  Then, 
1°.  We  must  obviously  have  R=0  in  {a'),  or 

P1+P2+P3  +  ,  &c.  =  0. 
2°.  If  this  value  of  R  be  substituted  in  (6'),  we  have 
__S.Pa_S.Pa; 
X~~BT~    0    * 


22  STATICS. 

Therefore  x  will  be  infinite  unless  2.Rr=0.  But  when  x  is  in. 
finite,  we  have  a  couple  {Art.  41),  and,  consequently,  there  can 
be  no  equilibrium.     Hence  we  must  also  have  2.P#=0,  or 

P1x1+P2.x2+P3a;3+,  &c.=0. 
And  in  the  same  manner  it  will  appear  that  2.P1/--O  in  case  of 
an  equilibrium.     But  2.P.r  and  S.Py  are  the  moments  of  the 
forces  in  reference  to  two  planes  parallel  to  their  directions. 
Hence  the  necessary  conditions  of  equilibrium  are, 

1°.   The  sum  of  the  forces  must  be  equal  to  zero. 
2°.   The  sum  of  their  moments,  with  reference  to  two  planes 
parallel  to  their  directions,  ?nust  each  be  equal  to  zero. 

48.  Cor.  1.  If  R=0,  but  S.Yx  and'S.Py  are  not  equal  to 
zero,  there  will  be  no  motion  of  translation,  but  simply  a  mo- 
tion of  rotation. 

For  if  Rj  be  the  resultant  of  all  the  positive  forces,  and  x , 
the  abscissa  of  its  point  of  application,  R2  the  resultant  of  all 
the  negative  forces,  and  x2  its  abscissa, 

then  Rj-z:,—  R2i2^0; 

or,  since  R,=R2,         Rl(x1—  £2);>0. 

But  Rj  being  finite,  x1—  x2  must  also  be  finite. 

That  is,  the  points  of  application  of  Rj  and  R2  are  not  the 
same,  and  a  tendency  to  motion  of  rotation  around  the  axis  of 
Y  exists. 

And  2.Py>0 

gives,  in  like  manner,  a  tendency  to  motion  around  the  axis 
of  a;. 

49.  Cor.  2.  If  the  equilibrium  subsist,  one  force,  as  Px,  must 
be  equal  and  opposite  to  the  resultant  of  all  the  others. 

50.  Cor.  3.  If  the  equilibrium  subsist  with  one  direction,  it 
will  subsist  whatever  be  the  common  direction  of  the  forces. 

51.    EXAMPLES. 

1.  Two  parallel  forces,  acting  in  the  same  direction,  have 
their  magnitudes  5  and  13,  and  their  points  of  application  A 


PARALLEL     FORCES.  23 

and  B  6  feet  distant.     Find  the  magnitude  of  their  resultant, 

and  its  point  of  application  G. 

Ans.  R=18, 
AC=4i 
BC=lf. 

2.  Find  the  resultant,  and  its  point  of  application,  when  the 
forces  in  the  last  question  act  in  opposite  directions. 

Ans.  R=8, 
AC=9f, 
BC=3f. 

3.  If  two  parallel  forces,  P  and  Q,  act  in  the  same  direction 

at  A  and  B,  and  make  an  angle  d  with  AB,  find  the  moment 

of  each  about  the  point  of  application  of  their  resultant. 

P.Q     „ 
Ans.  p—rrAB  sin.  d. 

4.  If  the  weights  1,  2,  3,  4,  and  5  lbs.  act  perpendicularly 
to  a  straight  line  at  the  respective  distances  of  1,  2,  3,  4,  and  5 
feet  from  one  extremity,  required  their  resultant  and  its  point 
of  application. 

Solution.  From  equations  (a)  and  (b)  we  have 

R=2.P=l  +  2+3+4+5=15  lbs. 

And,  taking  the  extremity  of  the  line  for  the  origin  of  co-or- 
dinates, 

R£=15.£=2.Pz=lXl+2X2+3X  3+4X4+5X5=55. 
.-.  5=3|  feet. 

5.  Let  the  weights  4,  —7,  8,  and  —3  lbs.  act  perpendicu- 
larly to  a  straight  line  at  the  points  A,  B,  C,  and  D,  so  that 
AB  =  5  feet,  BC=4  feet,  and  CD =2  feet;  find  the  resultant 
and  its  point  of  application  E. 

Ans.  R=2  lbs., 
AE=2  feet. 

6.  Let  three  forces  which,  if  concurring,  would  be  in  equi- 
librium, act  each  in  the  side  of  the  triangle  which  represents 
them  in  magnitude  and  direction.  Show  that  they  are  equiv- 
alent to  a  statical  couple. 


CHAPTER   III. 


THEORY     OF     COUPLES. 

52.  Def.  A  statical  couple  consists  of  two  equal  and  parallel 
forces  acting  in  opposite  directions  at  different  points  of  a  body. 

53.  Def.  The  arm  of  a  couple  is  the  perpendicular  distance 
between  the  directions  of  the  forces. 

54.  Def.  The  moment  of  a  couple  is  the  product  of  the  arm 
by  one  of  the  forces. 

55.  Prop.  A  couple  may  be  turned  round  in  any  manner  in 
its  own  plane  without  altering  its  statical  effect. 

Let  PjABP2  be  the  original  couple.     Suppose  the  arm  AB 

turned  around  any 
point  in  it  to  the  posi- 
tion ah.  Apply  the 
equal  and  opposite 
forces  P3  and  P4  per- 
pendicularly to  ab  at 
a,  and  similarly  P5 
and  P6  at  b,  and  let 
each  be  equal  to  Pj 
or  P2.  These  forces, 
being  in  equilibrium, 
will  not  disturb  the 
system.  The  two  equal 
forces  P,  at  A,  and  P4  at  a,  may  be  regarded  as  acting  at  their 
point  of  intersection  E,  and  will  have  a  resultant  in  the  direc- 
tion CE  bisecting  the  angle  P,EP4.  Similarly,  the  resultant 
of  P2  and  P6  at  D  will  be  an  equal  force  in  the  direction  CD. 
These  forces,  being  equal  and  opposite,  may  be  removed; 
that  is,  we  may  remove  from  the  system  the  forces  P,,  P2,  P4, 
and  P6,  and  we  have  remaining  the  forces  P3  and  P5  at  a 


COUPLES. 


25 


and  b,  forming  the  couple  P3a&P5,  which  is  the  same  as  if  we 
had  turned  the  original  couple  round  the  point  C  until  its  arm 
came  to  the  position  ab. 

56.  Prop.  A  couple  may  be  removed  to  any  position  in  its  own 
plane,  the  parallelism  of  its  arm  being  preserved,  without  chang- 
ing its  statical  effect. 

Let  the  arm  AB  of  the  couple  Pj  ABP2  be  removed  in  its 
own  plane  to  the  parallel  po- 
sition ab,  and  let  the  forces 
P3,  P4,  Ps,  P6,  each  equal  to 
the  original  forces,  be  applied 
perpendicularly  to  «&,  at  the 
extremities  a  and  b,  in  oppo- 
site pairs. 

Join  Kb  and  aB.  These 
lines  will  bisect  each  other 
in  C.  Pj  at  A  and  Ps  at  b 
will  have  a  resultant  2P,  at 
C,  parallel  to  the  original  direction.  Similarly,  P2  at  B  and 
P4  at  a  have  a  resultant  =  2P2  at  C,  opposite  to  the  former; 
these  will  consequently  balance  each  other,  and  may  therefore 
be  removed,  or  the  forces  P19  P2,  P4,  Ps  may  be  removed, 
&nd  we  have  remaining  the  couple  P3«&PG,  equivalent  to  the 
original  couple,  removed  parallel  to  itself  in  its  own  plane. 

57.  Prop.  A  couple  may  be  removed  into  any  other  plane,  par 
illel  to  the  original  one,  without  altering  its  statical  effect,  the 
parallelism  of  its  arm  being  preserved. 

Let  the  arm  AB  of  the  original  couple  be  transferred  from 
its  own  plane  MN  parallel  to  itself  to  ab  in  the  parallel  plane 
RS.  Let  forces  be  applied  at  a  and  b,  as  in  the  preceding 
proposition,  each  being  equal  to  Pa  or  P2.  Join  Kb  and  «B. 
These  lines  will  bisect  each  other  in  C.  The  forces  P,  and 
P6  will  have  a  resultant  —2?l  at  C,  and  P2  and  P4  a  result- 
ant =  2P2  at  C.  These  resultants  will  be  equal  and  opposite 
and  may  therefore  be  removed  without  disturbing  the  statica. 
effect  of  the  system.     We  have  then  remaining  the  couple 


26 


STATICS 


P3a&Ps,  equivalent  to  the  original  couple,  and  transferred  to 
the  plane  RS. 

58.  Prop.  All  statical  couples  are  equivalent  to  each  oilier 
whose  planes  are  parallel  and  moments  equal. 


Let  PjABP, 
0? 


E 


Q4 


AP2 


Q3 


y 


and  QjDEQ,2  be  any  two  couples  whose 
planes  are  parallel  and 
moments  equal.     If  they 

I>      are  not  in  the  same  plane, 

by  the  foregoing  propo- 
Yqj^  sitions  let  the  latter  be 
*  transferred  to  the  plane 

of  the  former,  and  moved 
in  that  plane  until  the 
extremities  D  and  A  of 
their  arms  coincide,  and 
DE   take    the    position 


Pl=P'+Ql 


AC.     Apply  at  C,  perpendicularly  to  AC,  two  equal  and  op- 
posite forces  Q3  and  Q4,  each  equal  to  QK     Resolve  P,  into 
two  forces  F  and  Qlf  so  that  P^P'+Q,,  or  P1-Q1=P/. 
Then,  by  hypothesis, 


COUPLES.  27 

P1.AB=Q1.DE=Q1.AC=Q1.AB.+Q1JBC=QrAB+Q3.BC 

and  (P1-Q1)AB=F.AB=Q3.BC, 

or  the  resultant  of  P'  and  Q3  (Art.  36)  passes  through  B.  It 
is  also  equal  and  opposite  to  P2.  Hence  the  forces  P',  Q,3, 
and  P2  may  be  removed,  and  there  will  remain  the  couple 
Q,ACQ4,  equivalent  to  PXABP2. 

59.  Prop.  Any  statical  couple  may  be  changed  into  another, 
which  shall  be  equivalent  and  have  an  arm  of  given  length. 

Let  PjPj  be  the  moment  of  any  couple,  of  which  Pj  is  one 
of  the  forces  and  p ,  its  arm,  and  let  p  be  the  given  arm.  Find 
a  fourth  proportional  P'j  to  p,  plf  and  P,  ;  or  take  p  :  px  = 
P,  :  P',,  which  gives  Y'\p=Y1p1.  Hence,  by  Art.  58,  their 
moments  being  equal,  and  in  the  same  plane,  the  couples  are 
equivalent. 

60.  Def.  The  aris  of  a  couple  is  a  line  perpendicular  to  the 
plane  of  the  couple. 

If  the  length  of  the  axis  be  taken  proportional  to  the  mo- 
ment of  the  couple,  and  drawn  above  its  plane  when  the  couple 
is  positive  or  tends  to  produce  rotatory  motion  in  the  direction 
of  the  hands  of  a  watch,  and  below  its  plane,  when  negative 
or  tends  to  produce  motion  in  the  opposite  direction,  then  the 
axis  will  completely  represent  the  couple  in  position,  intensity, 
and  sign. 

61.  Cor.  By  considering  the  previous  propositions,  it  will  be 
obvious  that  the  axis  of  a  couple,  as  thus  defined  and  limited, 
may  be  removed  parallel  to  itself,  to  any  position  within  the 
body  acted  on  by  the  couple. 

62.  Def.  The  resultant  of  two  or  more  couples  is  one  which 
will  produce  the  same  effect  singly  as  the  couples  themselves 
jointly. 

63.  Prop.  The  moment  of  the  resultant  of  two  or  more  couples 
in  the  same  or  parallel  planes,  equals  the  algebraic  sum  of  the 
moments  of  the  component  couples. 

Let  Pj,  P2,  P3,  &c,  be  the  forces  ;  plt  p2,  p3,  &c,  their 
arms  respectively.     The  couples  may  all  be  removed  into  one 


28 


t  a  r  i  c  s. 


Pi 


plane  (Art.  57),  turned  round  in  that 
plane  (Art.  55),  moved  in  it  (Art.  56), 
and  their  arms  changed  (Art.  59)  to  a 
common  arm,  while  their  moments  re- 
main the  same.  Let  p  be  the  com- 
mon arm  AB  ;  F,i  F2,  F3,  &c,  the 
forces;    so   that   V\p=Y^pl,  P' 2p= 

|pi    P2P2»   ?'zP=?zPv   &c-      Now    the 
forces  P'jjP^jFg  at  A  are  equiva- 

YEj    lent  to  a  force 


\(£ 


F1+F2+F3+,  &c.,=-L?-L+ 


2+^J?+,  &c. 


Vx 


p  p  P 

And  the  forces  at  B  are  equal  to  the  same  sum.     We  hav8 
then  for  the  moment  of  the  resultant  couple, 

(F ,  +Fa  +F  3 +,  &c.)  AB= (F  x  +Fa  +F ,  +)p, 

=PIp1+P3p2+P3p3+,&c. 

If   either   of   the   original 
couples, as  P3p3,tend  to  pro- 
p3    duce  motion  in  the  opposite 
A  'fl        direction,  its  sign  will  be  neg- 
ative.   The  sum  of  the  forces 
will  be 

P'j+F.-F.+.&a, 

and  the  moment  of  the  result- 
ant couple 
(F1+F2-F3+,&cOJp=P^1+P2jo2-P3/'i+,&c. 

64.  Cor.  The  axis  of  the  resultant  couple  will  be  equal  to 
the  algebraic  sum  of  the  axes  of  the  component  couples. 

Schol.  The  composition  of  couples  in  the  same  or  parallel 
planes,  by  means  of  their  axes,  is  therefore  analogous  to  the 
composition  of  conspiring  forces. 

65.  Prop.  To  find  the  resultant  of  two  couples  in  different 
planes  inclined  to  each  other. 

Let  each  couple  be  turned  round  and  moved  in  its  own 
plane  until  its  arm  coincide  with  the  common  intersection  of 
their  planes,  and  let  the  arms  be  made  coincident  and  of  the 


COUPLES. 


2<J 


same  length,  the  moments 
of  the  couples  remaining 
the  same.  Let  AB  be  this 
common  arm,  and  PABP, 
QABQ  the  couples.  Com- 
plete the  parallelograms 
on  the  lines  representing 
P  and  Q,  and  the  diago- 
nals will  represent  their 
resultants  R,  which  may 
be  substituted  for  them. 
Hence  we  have  the  result- 
ant couple  RABR  equiva- 
lent to  the  component  couples  PABP  and  QABQ. 

66.  Cor.  The  diagonal  of  the  parallelogram,  constructed  on 
the  lines  representing  the  axes  of  the  component  couples,  will 
represent  the  axis  of  a  couple  equivalent  to  them. 

Let  OK  and  OL  be  the 
axes  of  the  couples,  and  0  the 
angle  made  by  the  planes  of 
the  couples.  Since  the  axes 
are  perpendicular  to  these 
planes,  the  angle  made  by  the 
axes  will  be  6.  We  have, 
therefore,  0=PAQ=LOK. 

By  the  triangle  of  forces, 

R2=F+Q2+2PQ  cos.  0, 
.-.  R.AB 


ABVP2+Q2+2PQcos.6>, 


=  v'P.AB2+Q2.AB2+2P.ABxQ.ABcos.0: 
(since  OK=P.AB,  &c), 

=  v/OK2+OLs+20K.OL  cos.  0, 
=OM. 

And  OK,  OL  being  respectively  perpendicular  to  the  planes 
of  the  component  couples,  OM  will  be  perpendicular  to  the 
resultant  couple. 

67.  Schol.  The  above  is  analogous  to  the  parallelogram  of 
forces,  and  may  be  called  the  parallelogram  of  couples. 


80  STATICS. 

If  L  and  M  represent  the  axes  or  moments  of  the  compo- 
nent couples,  and  G  the  axis  or  moment  of  their  resultant,  then 
G2=L2+M2+2LM  cos.  6. 

If  L,  M,  N  were  the  axes  of  three  component  couples,  it 
might  be  shown  that  G,  the  axis  of  the  resultant  couple,  would 
be  represented  in  magnitude  by  the  diagonal  of  the  parallelo- 
piped  formed  upon  them. 

If  the  planes  of  the  three  couples  are  at  right  angles  each  to 
the  other  two,  or  if  L,  M,  and  N  are  at  right  angles  each  to 
the  planes  passing  through  the  other  two,  the  parallelopiped 
would  be  rectangular,  and  we  should  have, 
G2=L*+Ma+N\ 


CHAPTER  IV. 

ANALYTICAL    STATICS    IN    TWO    DIMENSIONS. 

In  this  chapter  the  points  of  application  and  the  directions 
of  the  forces  will  be  referred  to  co-ordinate  axes  at  right  an- 
gles to  each  other. 

68.  Prop.  To  find  the  magnitude  and  direction  of  the  result- 
ant  of  any  number  of  concurring  forces  in  the  same  plane. 

Let  Pj,  P2,  P3,  &c.  .  .  P„  be  the  n  forces,  and  let  the  point 


at  which  they  act  be  taken  for  the  origin  of  co-ordinates. 
Let  P,  make  the  angle  a,  with  OX,  and  /3X  with  OY. 
P2  "  a2  "  0a 

P3  «  a3  "  03         « 

&c,  &c,  &c, 

P„  "  an  "  A, 

If  OA  represent  the  force  P^and  the  parallelogram  OMAN 
be  completed,  OM  will  be  the  component  of  P,  in  OX,  and 
ON  that  in  OY.     And 


cos.  a j, 

Y1=P,  cos.  (3t, 

cos.  a2, 

Y2=P2  cos. /3a, 

cos.  as, 

Y3=P3  cos.  p3, 

&c, 

&c,         &c, 

;os.  a„, 

Y„=P„  cos.  |3B. 

32  STATICS. 

OM^X^P,  cos.  alt  ON=Y1=P1  cos.  0, 

Pursuing  the  same  course  with  the  forces  P2,  P3,  &c,  and 
calling  the  resolved  parts  of  the  forces  respectively  in  OX, 
X2,  X3,  X4,  &c.  .  .  .  X„, 

and  those  in  OY    Y2,  Y3,  Y4,  &c Yn, 

we  shall  obtain 

X1=P-, 
X2  =  P2 

X3  =  P>) 

&c, 
X„=Pn  c 

But  the  components  in  OX  are  equivalent  to  a  single  force, 

=X1+X1+X,+,  &c X„=2.X, 

and  those  in  OY  to  a  single  force, 

=Y1+Y2+Y3H-.  &c.  . .  .  Y„=2.Y, 
the  Greek  letter  2  being  used  as  the  sign  of  summation. 
Hence  we  have 

2.X=S.P  cos.  a,  (1) 

2.Y=S.P  cos.  (3.  (2) 

Now  if  R  be  the  resultant  required,  and  6  the  angle  it  makes 
with  OX,  the  resolved  parts  of  R  in  the  axes  must  equal  the 
resolved  parts  of  the  forces  in  the  same  directions. 

.-.  R  cos.  0=2.X,  (3) 

R  sin.  0=2.Y.  (4) 

2.Y 

Hence  Tan.  6= -p^-.  (5) 

Also,         R2.  cos.20+^sin.20=R2=(2.X)2+(Z.Y)2; 

or  R=V(2.X)2+(2.Y)2.  (6) 

Equations  (1)  and  (2)  give  the  values  of  2.X  and  2.Y,  by 
which  we  obtain  0  from  (5)  and  R  from  (6)  ;  or  R  may  be  ob- 
tained, numerically,  more  readily  from  the  equation 
R  cos.  0=2.X. 

For  R=-^r=2.Xsec.0.  (7) 

cos.  0 

69.  Schol.  It  is  readilv  seen  that  the  signs  of  the  compo- 


ANALYTICAL     STATICS. 


33 


nents  along  either  axis  are  involved  in  the  trigonometrical  ex- 
pressions of  their  values,  (1)  and 
(2).  Assume  the  directions  OX 
and  OY  positive,  and  OX',  OY' 
negative.  The  components  Ox ,, 
Oy1  of  Pj  are  positive ;  their 
values,  Pj  cos.  «,,  Pj  cos.  (3lt 
will  also  be  positive,  since  the 
cosines  of  angles  in  the  first 
quadrant  are  positive.  The  com- 
ponents of  P2,  which  lies  in  the 
angle  X'OY,  are  Ox2,  Oy2,  the 
former  negative  and  the  latter 
positive.  The  value  of  0.r2  = 
P2  cos.  a2  is  negative,  since  a2,  reckoned  from  OX  around  to 
the  left,  will  be  in  the  second  quadrant ;  and  that  of  Oj/2=P2 
cos.  (33  positive,  since  j32,  reckoned  from  OY  around  to  the 
right,  will  be  in  the  fourth  quadrant ;  or,  if  reckoned  from  OY 
around  to  the  left,  will  be  negative,  and  less  than  90°. 

The  components  Ox3,  Oy3  of  P3,  in  the  angle  X'OY',  are 
both  negative,  and  their  values  P3  cos.  a3,  P3  cos.  |33  are  neg- 
ative, because  the  angles  a3  and  (33  are  both  in  the  third  quad- 
rant. The  components  Ox4,  Oyi  of  P4,  which  lies  in  the 
angle  Y'OX,  are,  the  former  positive  and  the  latter  negative. 
The  value  of  0.r4=P4  cos.  a4  is  positive,  because  aA  is  in  the 
fourth  quadrant,  or,  if  taken  negatively,  is  less  than  90°  ;  that 
of  Oy4  =  P4  cos.  /34  is  negative,  because  /34  is  in  the  second 
quadrant.  By  the  above  mode  of  reckoning  the  angles,  we  al- 
ways find  a+/3=90°,  or  one  the  complement  of  the  other. 

70.  Prop.  Required  the  conditions  of  equilibrium  of  any 
number  of  concurring  forces  in  the  same  plane. 

Since  the  forces  are  in  equilibrium,  their  resultant  must  be 
equal  to  zero,  or  R=0.     This  gives,  by  (0), 

(2.X)S  +  (2.Y)2=0. 
But  each  term  being  a  square,  is  essentially  positive.     The 
equation,  therefore,  can  only  be  satisfied  by  making  each  term 
equal  to  zero  at  the  same  time. 

C 


34 


STATICS. 


Hence  2.X=0,  2.Y=0,  (8) 

are  the  two  necessary  and  sufficient  equations  of  equilibrium  of 
any  number  of  concurring  forces  in  the  same  plane ;  that  is, 
the  sums  of  the  components  of  all  the  forces  resolved  in  any 
two  rectangular  directions,  must  be  separately  equal  to  zero. 

71.  Prop.  Required  the  expressions  for  the  resultant  force 
and  resultant  couple  of  any  number  of  forces  acting  at  different 
points  in  the  same  plane. 

Let  P,,  P2,  &c.  .  .  .  Pn  be  any  number  offerees  in  the  same 
plane.  Take  the  co-ordinate  axes  OX,  OY  in  the  plane  of  the 
forces,  and  let  al,  a2,  a3  .  .  .  an  be  the  angles  the  forces  make 
respectively  with  OX,  and  (8,,  j32,  |33  .  .  .  j3n  the  angles  they 
make  with  OY. 

Let  s,,!/,  be  the 
co-ordinates  OM , , 
MjAj  of  A,,  the  point 
of  application  of  the 
force  Px,  and  x2.  y„, 
&c.  .  .  .  xn,  yn  be  the 
co-ordinates  of  the 
points  of  application 
of  the  others  respect- 
ively. Resolving  P, 
in  the  directions  of 
OX  and  OY,  we  shall 


Y^P,  cos.  j3lf 


have  for  the  components 

Xj— P,  cos.  a,, 
at  the  point  Aj. 

Now  apply  at  O,  in  OX,  two  equal  and  opposite  forces,  each 
equal  to  X,  ;  this  will  not  affect  the  system.  Then,  instead 
of  the  single  force  X,  at  Alf  we  have  Xl  at  A„  or  N,  and 
—X,  at  O,  which  together  form  a  couple,  and  Xj  at  0.  That 
is,  for  the  force  X,  at  A,,  we  may  substitute 

X,  at  O,  and  the  couple  X1.ON,=X1y1. 
In  like  manner,  applying  at  O,  in  OY,  two  forces,  opposite  and 
each  equal  to  Y x,  we  should  have  Y,  at  A,  equivalent  to 
Y,  at  O,  and  the  couple  -Y.OM1  =  -Y.a:1. 


ANALYTICAL     STATICS. 


35 


This  last  couple  will  be  negative,  since  it  tends  to  produce  ro- 
tation in  the  contrary  direction  to  that  of  the  other. 

By  pursuing  the  same  course  with  each  of  the  other  forces 
we  should  obviously  have  acting  at  O,  in  OX,  a  sum  of  forces 

X,+X2+X3+,&c X„=2.X,  (9) 

and  at  O,  in  OY, 

Y1+Y2+Y3+,&c Y„==2.Y,        (10) 

and  the  couples 

X12/1+X2y2+X33/3 X„ya=S.Xy,      (11) 

—Ylxl—Y2x2—Y3x3  ....  -Ynx,=2.  —  Yx.  (12) 
We  have  now  reduced  the  whole  to  a  system  of  concurring 
forces  and  a  system  of  couples.  '  The  couples  being  in  the 
same  plane,  the  moment  of-  the  resultant  {Art.  03)  will  be 
equal  to  the  algebraic  sum  of  the  moments  of  the  components, 
or  the  resultant  axis  G  will  be  equal  to  the  algebraic  sum  of 
the  component  axes.  Hence,  taking  the  sum  of  (11)  and  (12). 
G=2.Xy-2.Yx=X.(Ky-Yx).  (13) 

Now  if  R  be  the  resultant  force  acting  at  0,  and  6  the  angle 
it  makes  with  OX,  we  have,  as  before, 
R  cos.  0=2.X, 
R  sin.  0=2.Y. 

s.Y 

Tan.  0=S,  (14) 

and  R2=(2.X)'+(2.Y)2.  (15) 

72.  Schol.  1.  The  equations  (13),  (14),  and  (15)  determine 
the  magnitude  of  G  and  the  magnitude  and  direction  of  R. 
To  construct  these  results,  draw 
through  the  origin  O  the  line  OR, 
making  the  angle  0  with  OX,  to 
represent  R.  Then  put  G= 
£.(Xy— Yx)=Rr,  by  which  the 
moment  of  the  resultant  couple 
is  changed  into  another,  whose 
forces  are  each  equal  to  R,  and 
arm  equal  to  r.  Let  this  be 
moved  and  turned  round  until 
one  of  its  forces  acts  at  0  in  an 


36  STATICS. 

opposite  direction  to  the  resultant  force ;  draw  OA  perpendic 
ular  to  OR  and  equal  to  ;•,  and  AR'  parallel  to  OR.  Then 
R'AOR'  represents  the  resultant  couple.  The  two  forces  at  O. 
being  equal  and  opposite,  may  be  removed,  and  we  have  the 
final  resultant  acting  in  AR',  which  makes  the  angle  6  with 
OX. 

73.  Schol.  2.   To  find  the  equation  of  this  final  resultant.     It 
will  be  of  the  form 

y=ax+b. 

„    2.Y 

But  a=  tan.  6=- 


and  6=OB= 


2.X' 

OA  r  G  G 


cos.  d    cos.  d    R  cos.  d     2.X' 


Therefore,  by  substitution,  we  have  for  the  equation 

2-Y         G 

y=s3*+s3  (16)- 

74.  Prop.  To  determine  the  conditions  of  equilibrium  of  any 
number  of  forces  acting  at  different  points  in  the  same  plane. 

In  order  that  there  may  be  no  motion  of  translation,  we 
must  have  R=0,  which  gives,  as  before, 

2.X=0,  (17) 

and  £.Y=0.  (18) 

And,  in  order  that  there  may  be  no  motion  of  rotation,  we 
must  also  have 

G=l(Ky-Yx)=0.  (19) 

Equations  (17),  (18),  (19)  are  the  three  necessary  and  suffi- 
cient conditions  of  equilibrium. 

75.  Cor.  When  there  is  a  fixed  point  in  the  system,  if  this 
point  be  taken  for  the  origin,  its  resistance  will  destroy  the  ef- 
fect of  the  resultant  force  R,  and  the  sole  condition  of  equilib- 
rium will  then  be  G=0, 

or  l.(Xy-Yx)=0; 

or  there  must  be  no  tendency  to  rotation  around  the.  fixed 

point. 


ANALYTICAL     STATICS. 


37 


EQUILIBRIUM    OF    A    POINT    ON    A    PLANE    CURVE. 

76.  If  a  point  be  kept  at  rest  on  a  plane  curve  by  the  action 
of  any  number  of  forces  in  the  plane  of  the  curve,  the  resultant 
must  obviously  be  in  the  direction  of  the  normal  to  the  curve 
at  that  point,  and  equivalent  to  the  pressure  the  curve  sustains. 
For,  if  the  resultant  had  any  other  direction,  it  might  be  re- 
solved into  two,  one  in  the  direction  of  the  normal,  and  the 
other  in  the  direction  of  the  tangent  to  the  curve ;  the  former 
would  be  opposed  by  the  reaction  of  the  curve  ;  the  latter,  be- 
ing unopposed,  would  cause  the  point  to  move.     Hence, 

77.  Prop.  To  determine  the  conditions  of  equilibrium  of  a 
point,  retained  on  a  plane  curve  or  line,  by  forces  acting  in  its 
plane. 

Let  N  be  the  normal  force  of  reaction  of  the  curve,  and  a 
the  angle  made  by  the  normal  with  the  axis  of  x.  Also,  let 
2.X,  2.Y  be  the  sums  of  the  components  of  all  the  other  forces 
resolved  parallel  to  each  axis  respectively,  and  R  their  result- 
ant. The  resistance  N  may  be  considered  a  new  force,  which, 
together  with  the  other  forces,  retains  the  point  in  equilibrium 
independently  of  the  curve.  If,  therefore,  N  be  resolved  in 
the  direction  of  each  axis,  we  have  (8) 

N  cos.  a+2.X=0,  (a) 

N  sin.  a+2.Y=0;  (&) 

or,  transposing,  squaring,  adding,  and  reducing, 

N=  v'(2.X)2+(S.Y)2=R, 
or  the  reaction  is  equal  to  the  resultant  of  all  the  other  forces. 

Now  let  i  be  the  inclination 
to  the  axis  of  x,  of  the  tangent 
to  the  curve  through  the  point, 
then  a=90°  +  z. 


b.  T 


and 


.'.  cos.  a=cos.  (90°+i)  =  —  sin.  i, 
sin.  a=cos.i. 


Substituting  these  values  of  cos.  a  and  sin.  a  in  (a)  and  (b),  and 
dividing,  we  get 


38  STATICS. 

2  X 
Tan.  1=-™™  or  2.X+2.Y  tan.  i=0.      (20'. 

But  the  differential  expression  for  the  tangent  of  the  angl 

dxi 
which  the  tangent  makes  with  the  axis  of  a:  is  -£-.     Therefore. 
&  dx 

substituting  and  reducing, 

2.X.dz+2.Y.di/=0.  (21) 

Whenever  the  line  on  which  the  point  is  retained  is  right, 
(20)  may  be  used  ;  but  if  the  line  be  a  curve,  (21)  will,  in  gen- 

eral,  be  necessary,  and  -~  must  be  deduced  from  the  equation 

of  the  given  curve. 

VIRTUAL    VELOCITIES. 

78.  Def.  If  any  forces  Px  and  P2  act  at  the  point  A,  and 
this  point  be  displaced  through  an  in- 

1  definitely  small  space  Aa,  and  the 
perpendiculars  apx,  ap2  be  drawn 
from  a  on  the  directions  of  the  forces, 
then  Aj9j  and  Ap2  are  called  the  vir- 
tual velocities  of  the  forces  Pj  and  P2  ;  Aplf  measured  in  the 
direction  of  the  force  P15  is  positive,  and  Ap2,  measured  in  the 
direction  of  P2  produced,  is  negative. 

79.  The  principle  of  virtual  velocities  is  thus  enunciated : 

If  any  number  of  forces  be  in  equilibrium  at  one  or  more 
points  of  a  ~igid  body,  then  if  this  body  receive  an  indefinitely 
small  displacement,  the  algebraic  sum  of  the  products  of  each 
force  into  its  virtual  velocity  is  equal  to  zero. 

80.  Prof.  To  prove  the  principle  of  virtual  velocities  for 
concurring  forces  in  one  plane. 

Let  A  be  the  point  at  which  the  forces  P,,  P2,  P3,  &c 

P„  act ;  ax,  a2,  a3,  &c.  .  .  .  an  the  angles  they  make  respect- 
ively with  OX,  and  j3,,  j32,  (33  .  .  .  (3n  the  angles  they  make 
with  OY.  Let  0  be  the  angle  which  the  direction  of  the  dis- 
placement Aa  makes  with  OX 


ANALYTICAL     STATICS. 


sy 


■X 


Let  »,,  va,  v3,  &c.  .  .  .  vn  be  the  virtual  velocities  of  the 
forces  respectively. 
Then 
u1=Ajo1=Aacos.p1Aa=:Aa.  cos.  (a,— 0), 

=Aa(cos.a1  cos.  6+s'm.a1  sin.0), 
and  ¥lvl=l>l.Aa(cos.a1  cos.0-f  sin.aj  sin.0), 

= A«(cos.  0.P  ,  cos.  a ,  +sin.  0-Pj  sin.  a ,). 
In  like  manner,  for  P2  we  should  have 

P2u2  = Aa(cos.  0.P2  cos.  a2  +sin.  0.P2  sin.  a2), 
and  so  for  the  other  forces ;  and  taking  their  sum,  we  should 
get,  remembering  that  sin.  «=cos.  0, 

2.Pu=P1v1+P2v2+P3i>3+,  &c Vnvn, 

=  Aa[cos.  0  (P,  cos.  a,  +P2  cos.  a2  +,  &c P„  cos.  a„) 

+  sin.  0  (Pj  cos.  13,  +P2  cos.  /32  +,  &c P„  cos.  ft,)]. 

But  when  there  is  an  equilibrium  at  a  point  (8), 

Pj  cos.a,+P2  cos.a2+P3  cos.a3+,&c P„cos.a„—  2.X=0. 

and 

P,  cos.  |3,  +P2  cos./32  +P3  cos./33+,&c. . . .  P„  cos.ft  =2.Y=0. 

Hence  2.Pv=0, 

or  the  principle  is  true  when  the  forces  all  act  at  a  point. 

81.  Prop.  To  prove  the  principle  of  virtual  velocities  for 
forces  acting  at  different  points  in  the  same  plane. 

The  points  are  supposed  to  be  invariably  connected  by  rigid 
lines  or  rods  without  weight,  which  transmit  the  actions  and 
reactions  of  the  particles  or  points  upon  each  other. 


40  STATICS. 

Let  A,,  Aa,  A 3,  &c.  .  .  .  A„  be  the  particles  to  -which  the 

forces  Pj,  P2,  P3,  &c.  .  .  .  Pn  are  applied,  vx,  va,  v3  .  .  .  v„  the 

virtual  velocities  of  the  forces  respectively. 

Let  raaa  be  the  action  of  the  particle  Aj  upon  the  particle  A2, 

ra2a\       "      reaction  of         "        A2  "         "  A,, 

ra]at       "      action  of  "        A~  "         "  A3, 

ra^       "      reaction  of  "        A3  "         "  A,. 

&c,  &c,  &c. 

Let  Vaxa2,  va^,  Va^,  Va^,  &c,  &c,  be  the  corresponding 

virtual  velocities. 

Then  Va^^va^,  t7tf1o3=«asoi,  &c.,  &c,  from  the  nature  of 
action  and  reaction. 

Also,  raxa;t——raia1?  raiaa=—ra3ax,  &c,  &c.      For,  let  A, 

R'         A1P1 A2P2         J82 

M    '  ■  M-   . 


ax  a.2 

and  A2  be  the  particles  displaced  to  ax  and  a2.  Draw  the 
perpendiculars  axpl,  a2p2'  Then,  if  the  line  a xa2  is  par- 
allel to  AjA2,  it  is  obvious  that  A^p^=A2p2.  But  A,p,  is 
the  virtual  velocity  of  Rl5  and  A2j92  of  R2,  and  they  have  op- 
posite signs. 

.*.    Vaya=—Va2ax. 

If  ala2  is  not  parallel  to  A  XA2,  let  them  meet  when  pro- 

C— OT=~ Aijh Al[u- 

«»  a2 

duced  in  some  point  C.     Since  the  displacements  are  indefin- 
itely small,  the  perpendiculars  aJp1,a2p2  coincide  with  cir- 
cular arcs  whose  center  is  C,  and  Cal=Cp1,  Ca2=Cp2. 
But  A1^1=Cp1-CA1=Ca1-CA1> 

and  A„pa=Cpa—CAa=Caa—CA0  =  (Ca1+a1a0)-(CA1  + 

A.A,), 
=Gal—CAl. 
.:  AlPl  =    A2p2, 

Or  Valau=~"  Va^. 

Let  the  sum  of  the  products  of  all  the  forces  P,,  P2,  &c . 
into  their  virtual  velocities,  acting  on  the  particle 


ANALYTICAL     STATICS.  41 

A j  be  2.(Pai.va), 

those  on  A2  be  S.(Paa.raa), 

those  on  A3  be  2.(Pa3.Va3), 

&C,  &C, 

those  on  A„  be  I,.(PaH.vat). 

Since  each  particle  is  in  equilibrium  from  the  action  of  the 
external  forces  and  the  reactions  of  the  others  upon  it,  we 
have,  by  the  last  proposition, 

0  =  2.  (Pa^.Va)  +  1\a2.Vaia2  +  1'a^.Va^  + ,  &C, 
0=2.(Paa.Va^+ra;ia1.Vaaal+raaa3.Va3a3+i  &C, 
0  =  2.(Pa3.Vaa)+7'a3al.Va3al  +  raaaa.Vaaaa  +  ,  &C, 

&c,  &c,  &c, 

0  =  2.(P«>.Van)  +ra  Oj.r^fflj  +  ra  a2.Vona2  +  ,  &C. 

In  taking  the  sum  of  the  products  for  all  the  particles,  the 
products  of  the  reactions  into  their  virtual  velocities  will  dis- 
appear, being  in  pairs,  equal  in  magnitude  with  contrary 
signs  ;  therefore  we  have 

2.(Ya1.Va1)+2.{¥a2.Va2)+2.(Pa3.Va3)+,  &C 2.(PO|,.VOJ  =  0  : 

or,  generally,  when  there  is  an  equilibrium, 

S.(Pu)=0. 

82.  PRor.  Conversely.  If  the  sum  of  the  products  of  the 
forces  into  their  virtual  velocities  be  equal  to  zero,  or  2.(P.u.) 
=0,  then  there  will  be  an  equilibrium. 

For  if  the  forces  are  not  in  equilibrium,  they  will  be  equiva- 
lent either  to  a  single  force  or  a  single  couple  {Art.  74). 

In  the  first  case,  let  R  be  the  single  resultant  force ;  then  a 
force  equal  and  opposite  to  R  will  reduce  the  system  to  equi- 
librium ;  let  u  be  its  virtual  velocity  for  any  displacement. 
Since,  with  this  new  force,  there  will  be  an  equilibrium,  we 
have,  by  the  preceding  proposition, 

S.(P.»)+R.«=0. 

But  by  hypothesis,  2.(P.v)  =  0.    .'•  Ru=0,  which,  being  true  foi 
all  small  displacements  of  the  bodv,  we  must  have  R  =  0,  or 


42  STATICS. 

the  body  was  in  equilibrium  from  the  action  of  the  original 
forces. 

In  the  second  case,  if  the  forces  were  equal  to  a  resultant 
couple,  it  would  be  balanced  by  an  equal  and  opposite  couple. 
Let  the  forces  of  this  opposite  couple  be  Q,  and  Q',  and  their 
virtual  velocities  for  any  displacement  be  q  and  q'  respective- 
ly ;  since  they  will  reduce  the  system  to  equilibrium,  we  have, 
by  the  preceding  proposition, 

s.(P.v)+Q?+Qy=a 

But  2.(P.u)=0.  .-.  Qq+Q'q'=0,  for  all  displacements,  which 
is  impossible,  unless  Q,  and  Q/  each  equal  zero,  since  they  are 
parallel  forces,  and  act  at  difFerent  points. 


CHAPTER  V 


THE    CENTER     OF    GRAVITY. 


83.  Def.  Experiment  shows  that  ever ij  particle  of  matter  13 
subject  to  a  force  which  attracts  it  in  a  direction  perpendicu- 
lar to  a  horizontal  plane  or  the  surface  of  stagnant  water.  In 
reality,  the  directions  of  the  forces,  acting  on  several  particles, 
meet  nearly  in  the  center  of  the  earth ;  but  as  this  center  is 
very  distant,  compared  with  the  distance  of  any  particles  con- 
sidered together,  we  may,  without  sensible  error,  regard  their 
directions  as  parallel. 

This  force  is  cr.lled  gravity. 

84.  Experiment  shows,  also,  that  the  intensity  of  gravity 
varies  in  different  parts  of  the  earth's  surface ;  that  it  is  least 
at  the  equator,  and  increases  toward  the  poles  in  the  ratio  of 
the  square  of  the  sine  of  the  latitude.  It  shows,  also,  that  in 
the  same  latitude  the  intensity  varies  at  different  points  in  the 
same  vertical  line  ;  that  it  varies  inversely  as  the  square  of  the 
distance  from  the  center.  But  for  any  points  in  the  same  sys- 
tem, or  any  bodies  nearly  in  the  same  place,  this  variation  of 
intensity,  as  well  as  difference  of  direction,  may  be  neglected 
without  error.     Hence, 

85.  Def.  A  heavy  body  is  an  assemblage  of  material  points, 
or  particles,  acted  on  by  equal  parallel  forces  in  the  direction 
of  the  vertical  to  the  earth's  surface. 

86.  Def.  The  weight  of  a  body  is  the  resultant  of  all  the  ef- 
forts which  gravity  exerts  on  its  component  particles.  This 
resultant  {Art.  43)  is  equal  to  their  sum,  and  parallel  to  their 
common  direction.    • 

87.  Def.  The  mass  or  the  quantity  of  matter  of  a  body  is 
the  sum  of  all  its  component  particles. 

88.  Cor.  If  W  represent  the  weight  of  a  body,  M  the  mass, 
and  g  the  ratio  of  the  intensity  of  gravity  at  any  place,  to  its 


44  STATICS. 

intensity  at  another  place  where  it  is  assumed  as  unity,  \vc 
shall  have 

W=gM.  (22) 

For  the  resultant  of  all  the  parallel  actions  of  gravity  on  the 
particles  of  a  body  is  equal  to  their  sum,  or  the  product  of  its 
intensity  into  the  number. 

89.  Def.  The  density  of  a  body  is  the  ratio  of  its  mass  to  its 
volume  ;  or,  if  D  represent  the  density,  M  the  mass,  and  V  the 
volume, 

D=y;  (23) 

in  which  D,  M,  and  V  represent  the  number  of  units  of  each 
kind. 

90.  Cor.  Since  W=gM,  and  M=VD,  we  have 

W=gYD.  (24) 

91.  Prop.  The  masses  of  two  bodies  of  the  same  density  are  in 
the  direct  ratio  of  their  volumes. 

If  M,  V,  and  D  be  the  mass,  volume,  and  density  of  one 
body,  and  m,  v,  and  d  the  same  of  the  other,  by  (23), 

M=VD  and  m=vd;  v  M  :  m=VD  :  vd ;   (25) 
or,  since    D=d,  M  :  m=Y     :  v. 

92.  Prop.  The  masses  being  equal,  the  densities  are  inversely 
as  the  volumes. 

Since  M  :  m=VT>  :  vd;  if  M=m,  YD=vd; 

or  D  :  d=v  :  V.  (26) 

93.  Prop.  The  volumes  being  equal,  the  masses  are  directly  as 
the  densities. 

Making,  in  (25),  V=t>,  we  have  M  :  m=D  :  d.         (27) 

94.  Def.  The  center  of  gravity  of  a  body  is  the  center  of  the 
parallel  forces  of  gravity  on  each  of  its  component  particles. 

95.  Cor.  Hence  the  determination  of  the  center  of  gravity 
involves  an  immediate  application  of  the  doctrine  of  parallel 
forces,  and  we  need  only  refer  to  results  already  obtained  for 
many  important  deductions  respecting  the  center  of  gravity. 


CENTER     OF     GRAVITY.  45 

1°.  The  resultant  of  all  the  vertical  efforts  of  gravity  ou 
each  of  the  elementary  particles  of  a  body  passes  through  its 
center  of  gravity.     (Arts.  44  and  94.) 

2°.  This  resultant  is  parallel  to  the  forces  (Art.  36) ;  that  is, 
it  is  vertical,  and  its  magnitude  is  equal  to  the  weight  of  the 
body. 

3°.  Whatever  position  we  give  to  a  body,  this  resultant  will 
always  pass  through  the  center  of  gravity  ;  since  changing  the 
position  is  equivalent  to  changing  the  direction  of  the  forces, 
without  changing  their  points  of  application  or  parallelism. 

4°.  A  heavy  body  will  be  in  equilibrium  if  its  center  of 
gravity  be  supported,  whatever  may  be  the  situation  of  the 
body  relative  to  the  support,  since,  in  this  case,  the  resultant 
of  the  parallel  forces  of  gravity  will  have  a  fixed  point  in  its 
direction. 

5°.  When  we  wish  to  find  the  center  of  gravity  of  several 
bodies,  we  can  suppose  the  mass  of  each  concentrated  at  its 
center  of  gravity,  since  the  weight  of  each  is  a  force  propor- 
tional to  its  mass,  and  passing  vertically  through  its  center  of 
gravity.  Hence  we  have  only  to  consider  a  system  of  heavy 
points. 

96.  Def.  A  body  is  said  to  be  symmetrical  with  respect  to  a 
plane  when  the  lines  joining  its  particles,  two 
and  two,  are  parallel,  and  bisected  by  the 
plane. 

Thus,  let  m,  m'  be  two  symmetrical  par- 
ticles, so  that  the  line  mm'  may  be  bisected 
in  b  by  the  plane  aa' .  Letting  fall  the  perpen- 
diculars ma,  m'a',  the  equality  of  the  triangles 
mab,  m'a'b  gives  ma=m'a'.  Hence  the  par- 
ticles of  a  body,  symmetrical  with  respect  to 
a  plane,  are  situated,  two  and  two,  on  opposite  sides  of  the 
plane,  and  at  equal  distances  from  it. 

97.  Prop.  The  center  of  gravity  of  every  homogeneous  body 
of  uniform  density,  symmetrical  with  respect  to  a  plane,  is  situ- 
ated in  that  plane. 

For  any  two  particles,  symmetrically  placed,  will  be  at  the 


46 


STATICS. 


same  distance  from  the  plane,  and  their  moments  will  be  eqna 
and  have  contrary  signs.  But  all  the  particles,  taken  two  and 
two,  are  thus  placed  {Art.  96^.  Therefore  the  resultant  of  the 
system  of  forces  will  be  in  that  plane,  and,  consequently,  the 
center  of  gravity  also. 

98.  Def.  A  body  is  said  to  be  symmetrical  with  respect  to  an 
axis  when  it  is  symmetrical  with  respect  to  two  planes  pass- 
ing through  that  .axis. 

99.  Prop.  The  center  of  gravity  of  a  homogeneous  body,  sym- 
metrical with  respect  to  an  axis,  is  situated  in  that  axis. 

By  Art.  97,  it  must  be  in  each  plane  passing  through  the 
axis,  and  therefore  in  their  common  intersection,  or  the  axis 
itself. 

100.  Cor.  If  a  body  is  symmetrical  with  respect  to  two 
axes,  its  center  of  gravity  will  be  at  their  intersection,  since  it 
must  be  in  both  axes. 

101.  Def.  This  point  is  also  called  the  center  of  figure. 

102.  Prop.  To  find  the  center  of  gravity  of  any  number  of 
heavy  particles  whose  weights  and  positions  are  given. 

Let  A,  B,  C,  &c,  be  the  particles,  whose  weights  to,,  w2, 

wz,  &c,  act  at  their  respective 
centers  of  gravity,  vertically 
downward,  and  therefore  con- 
stitute a  system  of  parallel  for- 
ces. wx  and  w2  have  a  result- 
ant =w1+iv2  acting  at  some 
point  a,  such  that  iv1.Aa—iv2. 
Ba.  The  distance  AB  being 
given,  the  distanceA  a  is  de- 
termined by  taking  {Art.  39) 

w„ 
w,  +u)„  :  w0=AB  :  Aa=AB. -. — . 

Compounding  the  weight  wl+w2  at  a  with  another  weight 
w3  acting  at  C,  they  will  have  a  resultant  w1+w2-\-w3  acting 
at  some  point  b,  such  that 

(mj,  +w2).ab=w3.Cb. 


W3 


CENTER     OF     GRAVITY.  47 

First  determining  aC  from  the  triangle  aCB,  in  which  BC 
Ba,  and  the  angle  aBC  are  supposed  to  be  known,  we  can  de- 
termine the  distance  ab  by  the  proportion 

iv,  +w„  +U3,  :  w~=Ca  :  ab=Ca. ; • . 

1233  wl-\-w2+w3 

By  continuing  the  same  process,  we  should  determine  the 
point  at  which  the  final  resultant  weight  acts.  The  point  will 
be  the  same,  whatever  be  the  order  in  which  we  compound  the 

weights. 

103.  Prop.  To  find  the  center  of  gravity  of  any  number  of 
particles  in  the  same  plane  whose  positions  are  giver  by  their 
co-ordinates. 

Since  the  weights  of  the  particles  constitute  a  system  of  par- 
allel forces,  let  P,,  P2,  P3,  &c,  represent  the  weights  of  the 
particles  which  may  be  supposed  collected  in  their  respective 
centers  of  gravity,  x  , ,  yx,x2,y2,  &c,  their  co-ordinates.  We 
shall  have  for  the  co-ordinates  x,  y  of  the  center  of  parallel  for- 
ces (Art.  44),  or  center  of  gravity  of  the  whole  body  (Art.  94), 

_=P1x1+P2x0+P3.r3+,&c.^S.Px  ^ 
X  P1+P2+P3+,&c.  S.P     I 

-_P,y,+P8ya+P,y3+.&c-=s-Py  [   {  } 

y~       P,+P2+P3+,&c.  S.P'J 

104.  Cor.  1.  If  the  particles  all  lie  in  a  straight  line,  this  line 
maybe  taken  for  the  axis  of  x,  and  i/,  =0,  y2=0,  &c.  .*.  y=0, 
and  the  center  of  gravity  will  be  in  that  line. 

105.  Cor.  2.  If  the  particles  are  homogeneous,  the  weights 
of  each  particle  (24)  will  be  proportional  to  the  volumes ;  and 
if  «,,  v2,  u3,  &c,  denote  the  volumes  of  the  particles,  and  V 
the  whole  volume,  we  have 

X=~  ^^~  "^   I 

_       13,yi+132y2+133l/3+,  &C.       2.13?/     r 


y= 


i 


t) 


Hence  the  sum  of  all  the  particles,  or  the  whole  volume,  multi~ 
plied  by  the  distance  of  its  center  of  gravity  from  a  flane,  is 


48 


STATICS. 


equal  to  the  sum  of  each  particle  into  the  distance  of  its  cenle* 
of  gravity  from  the  plane. 

106.  Cor.  3.  If  the  center  of  gravity  of  the  whole  volume 
be  given,  and  the  center  of  gravity  of  one  of  its  parts,  the  cen- 
ter of  gravity  of  the  other  is  readily  obtained. 

Let  V  equal  the  whole  volume,  and  vlf  v2  the  volumes  of 
the  two  parts,  x,  y,  as,,  ylt  x2,  y2,  the  co-ordinates  of  their  re- 
spective centers  of  gravity. 


Then 


_      !),.£,  +V2X, 


,  and  y= 


»,yi+«9y3 


y        ' "~"  y  y 

But,  by  hypothesis,  V,  vl  and  x,y,xl,yl  are  given. 
Therefore,  u„=V — v., 


and 


Vx- 


■v.x 


-,  and  y. 


Vy-ViVi 


(30) 


A 


v2  ""  i>£ 

107.  We  shall  now  proceed  to  apply  the  foregoing  princi- 
ples to  specific  cases. 

Ex.  1.  To  find  the  center  of  gravity  of  a  uniform  physical 
straight  line. 

If  AB  be  the  uniform  straight  line,  and  C  its  middle  point,  C 

will  be  its  center  of  gravity  ; 
for  we  may  consider  the  line 
made  up  of  a  series  of  equal  particles  in  pairs  on  opposite 
sides  of  C,  and  the  weights  of  each  pair  would  be  equal  paral- 
lel forces  having  their  resultant  at  C,  the  middle  point  between 
them,  or  the  resultant  of  all  the  forces  will  pass  through  C. 

Or,  the  line  AB  is  symmetrical  with  respect  to  a  plane  pass- 
ing through  C.     The  center  of  gravity,  therefore  {Art.  97),  is  in 
jo  this  plane,  and  as  it  is  also  in  the  line 
AB,  it  must  be  at  their  intersection  C. 

Ex.  2.  To  find  the  center  of  gravity 
of  a  thin  triangular  plate  of  uniform 
density. 

Let  ABC  be  the  triangular  plate, 
of  which  the  thickness  is  inconsider- 
able. Bisect  AB  in  D,  and  AC  in  E. 
Join  C,  D,  and  B,  E  intersecting  in 


A 
G. 


G  will  be  the  center  of  gravity  of  the  plate. 


CENTER     OF     GRAVITY. 


49 


K 


Since  the  line  CD  bisects  all  lines  drawn  parallel  to  the 
base,  and,  consequently,  divides  the  triangle  symmetrically, 
the  center  of  gravity  of  the  triangle  will  be  in  this  line.  For 
the  same  reason,  it  will  be  in  the  line  BE,  and  will  therefore 
be  at  their  intersection  G. 

Join  D,  E.  DE  is  parallel  to  BC,  since  it  divides  the  sides 
AB  and  AC  proportionally,  and  DE=^BC. 

From  the  similar  triangles,  DEG  and  BCG,  we  have 
DE  :  BC=DG:GC  =  1  :  2.        .\  2DG=GC. 
Adding  DG  to  both  sides, 

3DG=DG+GC=DC.  .-.  DG=iDC. 

In  the  same  way  it  may  be  shown  that  EG=^EB. 

Hence  the  center  of  gravity  of  a  triangle  is  one  third  the 
distance  from  the  middle  of  either  side  to  the  opposite  vertex. 

Ex.  3.  To  find  the  center  of  gravity  of  a  parallelogram  of 
uniform  density  and  thickness. 

Bisect  the  sides  AB  and  DC  in  H  and  K  ;  also  the  sides 
AD  and  BC  in  E  and  F.  The 
plane  passing  through  H  and  K 
will  divide  the  parallelogram  sym- 
metrically, since  it  will  bisect  all 
lines  parallel  to  AB.  The  center 
of  gravity  will  lie  in  this  plane,  and 
will  therefore  lie  in  its  intersection  HK  with  the  parallelogram 
For  the  same  reason  it  will  lie  in  EF,  and  must  therefore  be  a! 
G,  the  common  point  of  these  lines.  B 

Or,  since  each  diagonal  bi- 
sects all  lines  drawn  parallel  to 
the  other,  it  will  be  at  the  inter- 
section of  the  diagonals. 

Ex.  4.  To  find  the  center  of 
gravity  of  a  thin,  polygonal 
plate,  of  uniform  density  and 
thickness. 

Let  ABCDEF  be  the  poly- 
gon. Draw  the  lines  AC,  AD, 
AE,  dividing  it  into  triangles. 

D 


H 


50  STATICS. 

When  the  polygon  is  given,  these  triangles  will  be  known, 
and  their  centers  of  gravity,  gx,g2,  g3,  gt,  may  be  found  by 
Ex.  2. 

The  mass  of  each  triangle  may  be  considered  as  a  heavy 
particle  at  its  center  of  gravity,  or  the  weight  of  each,  as  a 
force  acting  at  its  center  of  gravity.     Then  {Art.  39), 

gi+g2  -ga=giga  '-giG'^g,g2'Z~tr' 

6  1    '62 

This  determines  the  point  G',  the  center  of  gravity  of  the 
portion  ABCD.  In  the  same  manner,  we  may  find  the  point 
of  application  G",  of  the  resultant  of  gt+g2  acting  at  G',  and 
g3  acting  at  g3,  and  so  on;  the  last  point  so  determined  will 
be  the  center  of  gravity  G  of  the  polygon. 

Ex.  5.  To  find  the  center  of  gravity  of  a  triangular  pyramid 
of  uniform  density. 

Let  ABCD  be  the  triangular  pyramid.  Bisect  the  edge  BC 
in  E,  and  pass  a  plane  through  E  and  the  edge  AD.     This 

D 


plane  will  bisect  all  lines  in  the  pyramid  parallel  to  BC,  and 
will  therefore  divide  it  symmetrically.  Bisect  AC  in  F,  and 
the  plane  through  F  and  the  edge  BD  will  also  divide  the  pyr- 
amid symmetrically ;  and  since  the  center  of  gravity  of  the 
pyramid  will  be  in  both  these  planes,  it  must  be  in  their  inter- 
section D/.  But  the  point/ is  the  center  of  gravity  of  the  face 
ABC  {Ex   2).     Hence  the  center  of  gravity  of  the  pyramid 


CENTER     OF     GRAVITY. 


lies  in  the  line  drawn  from  a  vertex  to  the  center  of  gravity  of 
the  opposite  face.  Take  Ee=^ED,  and  join  Ae.  Since  e  is 
the  center  of  gravity  of  the  face  BCD,  the  center  of  gravity  of 
the  pyramid  will  be  in  the  line  Ae.  It  must  therefore  be  at 
the  intersection  G  of  Ae  and  D/l 

To  find  fG,  join  fe,  which  will  be  parallel  to  AD,  since  it 
divides  the  sides  ED  and  EA  of  the  triangle  AED  proportion- 
ally. Now  the  similar  triangles  fGe  and  AGD,  with/Ee  and 
AED,  give 

fG  :  GB=fe  :  AD=Ee  :  ED=1  :  3. 
.-.  3/G=GD. 
Adding  fG  to  both  members, 

4/G=/G+GD=  /D, 


or 


/G=i/D. 


Hence,  the  center  of  gravity  of  a  triangular  pyramid  is  one 
fourth  the  distance  from  the  center  of  gravity  of  one  face  to  the 
opposite  vertex. 

Ex.  6.  To  find  the  center  of  gravity  of  a  pyramid  whose  base 
is  any  polygon. 

In  the  pyramid  A  BCDEF  find  G,  the  center  of  gravity  of 
the  polygonal  base  (Ex.  4), 
and  join  AG.  Since  AG 
passes  through  the  center 
of  gravity  of  the  base,  it 
will  pass  through  the  cen- 
ter of  gravity  of  every  sec- 
tion parallel  to  the  base, 
and  the  center  of  gravity 
of  the  whole  pyramid  will 
be  in  AG. 

Join  GB,  GC,  GD,  GE,  b< 
and  GF,  and  conceive 
planes  to  pass  through  A 
and  each  of  these  lines, 
thus  dividing  the  whole 
pyramid  into  as  many  triangular  pyramids  as  the  base  has 


52  STATICS. 

sides.  The  centers  of  gravity  of  these  pyramids  will  be  at 
one  fourth  the  distances,  respectively,  from  the  centers  of  grav- 
ity of  their  triangular  bases  to  the  common  vertex  A.  These 
distances  being  thus  divided  proportionally,  the  points  of  di- 
vision will  all  lie  in  the  same  plane  parallel  to  the  base.  And 
since  the  centers  of  gravity  of  all  the  triangular  pyramids  are 
in  this  plane,  the  center  of  gravity  of  the  whole  pyramid  will 
be  in  it,  and,  being  in  the  line  AG  also,  will  be  at  their  inter- 
section g.  But  the  plane  divides  all  lines  drawn  from  the  ver- 
tex A  to  the  base  proportionally ;  therefore,  the  center  of  grav- 
ity g  of  the  whole  pyramid  is  one  fourth  the  distance  from  the 
center  of  gravity  of  the  base  to  the  vertex. 

Cor.  Since  the  above  principle  is  true,  whatever  be  the  num 
ber  of  sides  of  the  polygon,  it  is  true  when  the  number  becomes 
indefinitely  great,  or  when  the  base  becomes  a  continued  closed 
curve,  as  a  circle,  an  ellipse,  &c. ;  or,  the  center  of  gravity  of  a 
cone,  right  or  oblique,  and  on  any  base,  is  one  fourth  the  dis- 
tance from  the  center  of  gravity  of  the  base  to  the  vertex. 

Ex.  1.  To  find  the  center  of  gravity  of  a  frustum  of  a  cone  or 
pyramid  cut  off  by  a  plane  parallel  to  the  base. 

Let  a  be  the  length  of  the  line  drawn  from  the  vertex  of  the 
cone,  when  complete,  to  the  center  of  gravity  of  the  base,  a' 
that  portion  of  it  between  the  vertex  and  the  smaller  base  of 
the  frustum.     Then  (30)  we  have 

vx—v,x, 


in  which  x—^a,  xx=\a'.  Now  the  part  of  the  cone  or  pyra- 
mid cut  off  is  similar  to  the  whole,  and  similar  solids  are  as  the 
cubes  of  their  homologous  dimensions,  or  cubes  of  their  lines 
similarly  situated. 

Hence  v  :  u,=a3  :  a'\  or  v—vx  :  v=as—a"  :  a3. 


a" 


and  va  =  v—v1=v[l -J, 


CENTER     OF     GRAVITY. 


53 


__*fr-pfe)*,'_aq'-q''      , 
*  X"~        (       a/3\ 


a  —a' 


4    a2+aa'+a"    ' 

Subtracting  this  from  a,  we  have  the  distance  of  the  center 
of  gravity  of  the  frustum  from  the  center  of  gravity  of  its  base 
equal  to 


3a'3 


a—i- 


,(a+a')(ai+an)_a 

cf+aa'+a'*    ~4~ '  4(a*+aa'+an)' 


O 


X 


Ex.  8.   To  find  the  center  of  gravity  of  the  perimeter  of  a  tri- 
angle in  terms  of  the  co-ordinates   y 
of  the  angular  points. 

In  the  triangle  ABC,  let  a,  b, 
c  represent  the  sides  respective- 
ly opposite  to  the  angles  A,  B, 
C.  Their  centers  of  gravity  will 
be  each  at  the  middle  point  of 
the  side;  as  glt  g2,  g3. 

Let  xlyl  be  the  co-ordinates  of  A  referred  to  the  origin  O, 
x2y2  "         "         "  B, 

x3y3  «         "•        »  C. 

Then  the  co-ordinates  of  gz  are  ^(x^+x^,  \(V  1+2/2)' 

g2    »    \{x2+x~3),\{y2+y3), 
g3    «   ^(x3+x,),\{y3+yi), 

and  x,  y,  being  the  co-ordinates  required,  by  (28),  we  have 

a(x2  +x3)+b(xl  +x3)+c(x1  +x2) 
2(a+b+c) 
-__^a{y2+y2)+b(y,+y3)-\-c{y,+y2) 
y  2(a+b+c) 

Ex.  9.  To  find  the  co-ordinates  of  the  center  of  gravity  of  a 
triangle. 

Let  xxy^,  x2y2,  x3y3  be  the  co-ordinates  of  the  points  A 


x=- 


54 


STATICS. 


O     L 


or 
and 


N        M 


B,  C  respectively.  Draw 
AD  bisecting  BC  in  D, 
and  take  AG=|AD :  G 
is  the  center  of  gravity 
of  the  triangle.  The  co- 
ordinates of  D  are  ^(x2 

+a;3)»i(ya+y3);  and  if 

£=ON,  y=GN,  be  the 
co-ordinates  of  G,  we 
have 


ON=OL+f(OM-OL), 

GN=AL+f (DM-AL) ; 

y=yi+%{Uy2+y3)-yA=Wji+y2+y3)' 


CONDITIONS    OF    EQUILIBRIUM    OF    BODIES    FROM    THE    ACTION    OF 

GRAVITY. 

108.  Prop.  If  a  body  have  a  fixed  point  in  it,  the  condition  of 
equilibrium  requires  that  the  vertical  line  through  the  center  of 
gravity  shall  pass  through  the  fixed  point. 

If  the  center  of  gravity  g  be  in  the  vertical  line  Ag,  passing 

through  the  fixed  point 
A,  the  weight  w  of  the 
body,  being  a  vertical 
force  acting  at  g,  in  the 
direction  Ag,  will  be 
resisted  by  the  reaction 
of  the  fixed  point  A. 
If  the  center  of  gravity 
be  at  any  other  point 
g',  then  drawing  the 
vertical  line  through  g* 
and  the  horizontal  line  through  A,  the  weight  w  acting  at  g' 
would  have  an  uncompensated  moment  w.Am,  which  will  not 
vanish  until  the  center  of  gravity  comes  into  the  line  Ag.  Or, 
if  g'  be  the  center  of  gravity,  the  body  will  be  acted  upon  by 


CENTER     OF     GRAVITY. 


a  couple  of  which  the  forces  are,  the  weight  of  the  body  at  g', 
and  the  reaction  of  the  fixed  point  Alf  and  the  arm  Am. 
Therefore  (19),  in  order  to  equilibrium, 
S(Xy— Yx)=w.Am=0. 

.'.  A?n=0,  or  the  point  g'  must  be  in  Ag. 

109.  Def.  The  equilibrium  is  said  to  be  stable  when  the  body, 
if  slightly  disturbed,  tends  to  return  to  its   original  position. 
It  is  called  unstable   when,  being  disturbed,  it   tends   to  re 
move  further  from  its  original  position  ;  and  neutral  when,  after 
being  disturbed,  it  still  remains  in  equilibrium. 

110.  Prop.  When  the  equilibrium  of  a  body  containing  a 
fixed  point  is  stable,  the  center  of  gravity  is  in  the  lowest  posi- 
tion it  can  take ;  when  unstable,  in  the  highest. 

Let  A  be  a  fixed  point  in  the  bodies  M  and  N,  g  their  ren- 
ter of  gravity.  The 
center  of  gravity  can 
only  move  on  the 
surface  of  a  sphere 
whose  center  is  A 
and  radius  Ag.  When 
the  body  M  is  dis- 
turbed, its  center  of 
gravity  g  being  re- 
moved to  g',  will  rise 
through  the  versed  sine  of  the  arc  gg',  and  when  at  g',  the 
moment  iv.Am  will  obviously  tend  to  bring  it  back  to  its  orig- 
inal position.  The  equilibrium  is  therefore  stable,  and  the 
center  of  gravity  the  lowest  possible.  In  the  body  N,  the  cen- 
ter of  gravity  being  at  g  is  the  highest  possible,  and  being  in 
the  vertical  Ag-,  will  be  in  equilibrium.  When  removed  to  g', 
the  moment  w.Am  will  obviously  tend  to  carry  it  further,  and 
the  equilibrium  was  therefore  unstable. 

111.  Cor.  1.  The  pressure  on  the  point  by  which  a  body  is 
suspended  is  clearly,  in  the  case  of  equilibrium,  equal  to  the 
weight  of  the  body. 

112. -Cor.  2.  If  a  body  is  suspended  from  two  points,  the  po- 
sition of  equilibrium  is  that  in  which  the  center  of  gravity  is  in 


56 


STATICS. 


the  vertical  plane  passing  through  the  two  points  of  suspem 
sion,  since  it  is  then  the  highest  or  low 
est  possible.  To  determine  the  pres- 
sures on  the  fixed  points  A  and  B,  let 
GC  represent  the  weight  of  the  body 
acting  vertically  at  G.  Resolve  GC 
into  the  two  forces  DG  and  EG  acting 
in  the  directions  AG  and  BG.  These 
will  represent  the  pressures  on  A  and 
B.     Since  the  directions  of  GC,  GD, 

and  GE,  and  the  magnitude  of  GC  are  known,  the  magnitudes 

of  DG  and  EG  may  be  determined. 

113.  Cor.  3.  If  a  body  be  suspended  from  three  fixed  points 
not  in  a  right  line,  the  body  is  necessarily  at  rest. 

With  regard  to  the  pressures  on  each  point,  the  three  lines 
drawn  from  the  fixed  points  to  the  center  of  gravity  give  the 
directions  of  the  pressures,  and  the  vertical  is  the  direction  of 
the  weight,  the  magnitude  of  which  is  given.  Hence  we  have, 
in  a  parallelopiped,  the  three  sides  and  diagonal  given  in  po- 
sition and  one  given  in  magnitude,  to  determine  the  magnitude 
of  the  other  three. 

114.  Cor.  4.  If  a  body  touch  a  horizontal  plane  in  one  point, 
it  will  be  in  equilibrium  when  the  vertical  through  its  center 
of  gravity,  and  the  perpendicular  to  the  plane  at  the  point  of 
contact,  coincide  ;  for  the  weight  will  then  be  counteracted  by 
the  plane. 

115.  Cor.  5.  If  the  body  touch  the  plane  in  two  points,  it 
will  be  in  equilibrium  when  the  verti- 
cal through  the  center  of  gravity,  and 
the  perpendiculars  to  the  plane  at  the 
points,  are  in  the  same  plane.  Thus, 
if  ABC  be  a  vertical  section  through 
the  two  points  of  support  P  and  Q, 
there  will  not  be  an  equilibrium  unless 
the  vertical  through  the  center  of  grav- 
ity G  is  in  the  same  plane,  or,  which  is  the  same  thing,  meets 


CENTEROPGRAVITY.  57 

the  line  PQ.     The  pressures  on  P  and  Q  may  be  determined 
by  the  theory  of  parallel  forces,  and 

P  :  Q  :  W=WQ  :  WP  :  PQ, 

P,  Q,  and  W  representing  the  pressures  on  the  two  points  and 
the  weight  of  the  body  respectively. 

If  the  body  touch  the  line  PQ,  in  more  than  two  points,  the 
problem  of  the  pressures  is  indeterminate,  as  the  pressures 
may  be  any  how  distributed. 

116.  Cor.  6.  If  the  body  touch  the  plane  in  three  points,  it 
will  be  in  equilibrium  when  the  vertical  through  the  center  of 
gravity  falls  within  the  triangle  formed  by  joining  these  points. 

To  estimate  the  pressures  in  this  case,  let  PQR  be  the  tri- 
angle formed  by  joining  the  three 
points,  and  W  the  point  where  the 
vertical  through  the  center  of 
gravity  meets  it.  We  must  re- 
solve the  weight  acting  at  W  into 
three  others  parallel  to  it,  acting  at 
P,  Q,  and  R.  Join  PW,  QW,  and 
RW,  and  produce  them  to  the  op- 
posite sides.     Then,  by  the  theory  of  parallel  forces, 

P  :  W=WK  :  PK= triangle  WRQ  :  triangle  PRQ, 
Q  :  W=WM  :  QM=       "        WRP  :      "  "     PRQ, 

and         R:W=WH:RH=      "        WPQ:      "        PRQ. 

\  P:Q:R  :  W=WRQ  :  WRP  :  WPQ  :  PQR. 

If  the  body  touches  the  plane  in  more  than  three  points,  the 
pressures  on  the  points  are  indeterminate,  but  their  sum  is 
equal  to  the  weight  of  the  body. 

117.  Cou.  7.  If  the  vertical  through  the  center  of  gravity  of 
a  body  on  a  plane  meets  the  plane  in  a  point  within  the  base, 
the  body  will  stand.  For  the  resultant  of  the  parallel  forces 
of  resistance  must  be  within  the  figure  formed  by  joining  the 
several  points  of  contact. 

If  the  vertical  falls  without  the  base,  we  have  two  parallel 
forces  in  contrary,  but  not  opposite,  directions,  and  the  body 
will  turn  over. 


58 


STATICS. 


118.  Prop.  The  stability  of  a  body  is  measured  by  the  excess 
of  the  shortest  line  that  can  be  drawn  from  the  center  of  gravity 
to  the  perimeter  of  the  base,  above  the  vertical,  from  the  center  to 
the  horizontal  plane. 

The  stability  will  depend  on  the  ex- 
cess of  GP  over  GH,  since  G  must  be 
elevated  a  distance  equal  to  this  dif- 
ference, in  order  to  turn  the  body  over 
the  edge  of  the  base  at  P. 


P       H 

Cor.  1.  The  greater  the  base  HP  the  greater  the  stability, 
if  the  height  of  G  remain  the  same  ;  and  the  greater  HG,  the 
less  the  stability  if  HP  remain  the  same. 

Cor.  2.  The  stability  is  measured  by  the  versed  sine  of  the 
arc,  through  which  the  center  of  gravity  must  move  from  rest 
to  its  highest  point.  For  GP-GH=GP-BP=AP-BP=AB 
—  versed  sine  of  arc  GA  to  radius  PG. 

119.  Prop.  If  a  body  be  placed  on  an  inclined  plane,  it  will 
descend  when  there  is  no  resistance  from  friction. 

For  the  weight  of  the  body,  represented  by  GA,  may  be  re- 
solved into  two  others,  GB 
and  BA,'  one  perpendicular 
to  the  plane,  and  the  other 
parallel  to  it,  of  which  GB, 
the  one  perpendicular  to  the 
plane,  can  alone  be  counter- 
acted by  the  plane.  In  this 
case,  if  the  vertical  GA  fall  within  the  base  of  the  body,  the 
body  will  slide  ;  if  it  fall  without,  it  will  slide  also. 


120.    EXAMPLES. 

1.  If  two  right  cones  have  the  same  base  and  their  vertices 
in  the  same  direction,  find  the  distance  of  the  center  of  gravity 
of  the  solid  contained  between  their  two  surfaces  from  their 
common  base. 

Ans.  \  sum  of  their  altitudes 


CENTER     OF     GRAVITY.  59 

2.  The  center  of  gravity  of  a  paraboloid  being  in  the  axis 
at  a  distance  from  the  vertex  equal  to  §  of  the  axis  ;  fina  the 
center  of  gravity  of  a  frustum  of  a  paraboloid  from  the  base, 
a  and  b  being  the  radii  of  the  two  ends,  and  m  the  parameter 
to  the  axis  (30). 

a6-3aibi+2bB 
AUS'      3//i(a4-i4)    • 

3.  Two  spheres,  whose  radii  are  a  and  b,  touch  each  othei 

internally;  find  the  distance  of  the  center  of  gravity  of  the  solid 

contained  between  the  two  surfaces  from  the  point  of  contact. 

a'+a?b+ab*+b3 

Ans. 5 — t- — . 

a'+ab+b2 

4.  The  distance  of  the  center  of  gravity  of  a  hemisphere 
from  its  base  being  §  the  radius,  find  that  of  a  hemispherical 
bowl  whose  internal  radius  is  a  and  thickness  c. 

3  4a3  +  6a"c+4ac"+c3 
AnS-  ~6'       3a2+3ac+c2       '" 

5.  From  the  result  obtained  in  Ex.  4,  find  the  distance  of  the 
center  of  gravity  of  a  hemispherical  surface  from  the  center 
of  the  base. 

Ans.  \a. 

APPLICATION    OF   THE    PRINCIPLES    OF   THE    INTEGRAL   CALCULUS    TO 
THE    DETERMINATION    OF    THE    CENTER    OF    GRAVITY. 

121.  By  the  principles  of  the  integral  calculus,  when  the 
volumes  v  {Art.  105)  become  indefinitely  small,  they  may  be  re- 
garded as  the  differential  elements  of  the  body,  and  be  repre- 
sented by  do.     In  this  case  formulas  (29)  will  take  the  form 

fxdv      _    fydv 

7&T'   y==JdV' 

in  which  x  and  y  denote  the  distances  of  the  center  of  gravity 
of  dv  from  the  co-ordinate  axes. 

122.  Prop.  Required  the  differential  expressions  for  the  co-or- 
dinates of  the  center  of  gravity  of  a  plane  curve  or  line. 

If  ds  represent  the  differential  element  of  the  curve  or  line, 
by  substituting  ds  for  dv  in  (31),  we  have 


%F=-Z£r>    y=-7vAT'  (31> 


60  STATICS. 

_    fxds     _    fyds 

*=— >  y=— ■  (32) 

If  the  arc  is  symmetrical  with  respect  to  the  axis  of  xy  the 
center  of  gravity  will  be  in  that  axis  {Art.  99),  and  y=0. 

_    fxds 
s 

s  sufficient. 

123.  Prop.  Required  the  differential  expressions  for  the  cu~ 
ordinates  of  the  center  of  gravity  of  a  plane  area. 

Since  the  differential  element  of  a  plane  area  is  dxdy,  dv= 
dxdy.     By  substitution  in  (31),  we  have 

-Jfxdxdy  _=ffydxdy 
ffdxdy  '  y     ffdxdy  ' 
Integrating  in  reference  to  y,  we  have 

^_fxydx  -JMte  m 


fydx  '  y       fydx 
If  the  area  is  symmetrical  with  respect  to  the  axis  of  x,  the 
center  of  gravity  is  in  that  line  {Art.  99),  and  y—0. 

_    fxydx 

' '  x~  fvdx 

is  sufficient. 

124.  Prop.  Required  the  differential  expressions  for  the  co- 
ordinates of  the  center  of  gravity  of  a  surface  of  revolution 
around  the  axis  of  x. 

The  center  of  gravity  will  obviously  be  in  the  axis  of  x,  and 
therefore  ^=0;  and  since,  for  a  surface  of  revolution,  dv= 
2-nyds,  the  first  of  equations  (31)  become, 

-    fxyds  <ia\ 

*=fyte>  (34) 

and  this  equation  is  sufficient. 

125.  Prop.  Required  the  differential  expressions  for  the  co* 
ordinates  of  the  center  of  gravity  of  a  solid  of  revolution. 

In  this  case  dv=-nifdx.  Hence,  from  the  first  of  equations 
(31),  we  have 


•  CENTER     OF     GRAVITY.  61 

_     fy^xdx 

'r~7pzp  (35) 

which  alone  is  sufficient. 

By  proper  substitutions  for  do  in  the  fundairental  equations 
(31),  we  may  find  expressions  for  the  co-ordinates  of  the  cen- 
ter of  gravity  for  other  forms  of  bodies. 

126.  Prop.  The  surface  generated  by  the  revolution  of  a  curve 
around  an  axis  is  equal  to  the  length  of  the  curve,  multiplied  by 
the  circumference  described  by  its  center  of  gravity. 

From  the  second  of  equations  (32),  we  have 

2TT.y.s=2rrfyds. 

Now  2rry  is  the  circumference  of  which  y  is  the  radius,  and 
2ny.s  is  the  circumference  described  by  the  center  of  gravity 
of  the  curve  s  in  its  revolution  round  the  axis  of  x,  multiplied 
by  the  length  of  the  curve  s.  But  this  is  equal  to  2nfyds,  which 
is  the  area  of  the  surface  generated  by  the  revolution  of  the 
curve.     Hence,  &c, 

127.  Prop.  The  volume  generated  by  the  revolution  of  a  plane- 
area  around  an  axis  is  equal  to  the  product  of  that  area  by  the 
circumference  described  by  its  center  of  gravity. 

For,  from  the  second  of  equations  (33),  we  have 

2nyfydx = -nfy^dx. 

In  this  equation,  fydx  is  the  generating  area,  2iry  is  the  cir- 
cumference described  by  its  center  of  gravity,  and  -nfifdx  is 
the  volume  generated.     Hence  the  truth  of  the  proposition. 

These  last  two  propositions  comprise  the  theorem  otGuldin, 
and  their  application  to  the  determination  of  the  surfaces  and 
volumes  of  bodies  constitutes  the  Centrobaryc  Method.  By 
this  method,  of  the  three  quantities,  viz.,  the  generatrix,  the  dis- 
tance of  the  center  of  gravity  from  the  axis,  and  the  magnitude 
generated,  any  two  being  given,  the  other  may  be  determined. 


62 


STATICS. 


128.    EXAMPLES. 

1.  Required  the  center  of  gravity  of  a  circular  arc. 

Let  the  axis  of  a:  bisect  the  arc  MAM 

in  A,  the  origin  being  at  the  center  of 

the  circle,  and  let  MAM' =25.     From  the 

x  equation  of  the  circle  ?/2=r2— a;2,  we  ob- 

.     dy2        x2 
tain  -~= 


dy2 


dx2     r2—x2 
rdx 


But  ds=dx\/  1-f 


dx2        Vr-f 
fxds         r  P    xdx 


ry 


Vr'-x* 


+-Vr2-x2+C=+^+C. 
s  s 


ry 


When  y—0,  x=r,  and  —  =r.     .'.  C=0 

Hence  *=CI=^=^; 

s        2s   ' 

or,  the  distance  of  the  center  of  gravity  of  a  circular  arc  from 

the  center  of  the  circle  is  a  fourth  proportional  to  the  arc,  the 

radius,  and  the  cord  of  the  arc. 

If  the  arc  be  a  semicircle,  y=r,  and  s=^7rr 

2r 
.'.  CI=— =0.63662/-. 

2.  Required  the  center  of  gravity  of  a  circular  segment. 
Putting  CP=a  {Fig.,  Ex.  1),  and  taking  the  center  for  the 
origin,  we  have  y=  Vr2—x2.     Hence  (33), 

^_f:xydx_f:x{r2-x2)hxJt{r2-a2Y-JsW? 
X~  f'aydx~         MAP        ~    MAP    "MAP' 
__?V(2MP)8_Ty(chord)3 
"""  ^--^MAM'P"  segment  * 


If  the  segment  is  a  semicircle, 

_     TV  (2r)3     4r 

a:=    V    /  =—  =  0.42441r: 

T>7TV  3T 


ici. 


CENTER     OF     GRAVITY. 


G3 


3.  Required  the  center  of  gravity  of  the  surface  of  a  spher- 
ical segment. 

Taking  the  origin  at  C  (Fig.,  Ex.  1),  the  center  of  the  gen- 

dy2    x1 
erating  circle,  we  have  x*+y2=r\  j^=—,  and  yds=rdx. 


(34)  x-- 


dx2 
fyxds     fxdx 


fyds       fdx ' 
Integrating  between  x=r  and  x=a=CP. 


r—a 


Hence  the  center  of  gravity  is  at  the  middle  of  PA. 

4.  Required  the  Center  of  gravity  of  a  spherical  segment. 

Taking  the  origin  at  A,  the  vertex  of  the  generating  circle, 
we  have,  for  its  equation,  yi=2ax—xi. 


.-.  (35)  x= 
Hence       AG= 


fxy^dx     f  (2ax — z*)  xdx 
fy'dx       f(2ax—x~)dx  ' 
%ax3— \x*     8  ax— 3a:2 


3x 


12a— 4x' 


If  the  segment  is  a  hemisphere,  x=a, 
and  x—  fa. 

5.  Required  the  surface  of  a  hemisphere. 

By  the  centrobaryc  method  (Art.  126),  we  have 

The  generatrix  =^Ttr,  the  ordinate  of  its  center  of  gravity 

2r 
y= —  (Art.  128,  Ex.  1),  and  the  circumference  described  by 

77 

2r 
the  center  of  gravity  =27t.—  =  4r.    Hence  the  surface  =\irrAr 

=2nr\ 


129.    EXAMPLES    ON    THE    PRECEDING    CHAPTERS. 

Ex.  1.  Two  beams,  rigidly  connected  at  a  given  angle,  turn 
on  a  horizontal  axis  through  their  point  of  union  ;  find  the  po- 
sition of  equilibrium  by  the  action  of  their  own  weights. 

Let  AC,  BC  be  the  beams  suspended  from  C,  and  makino 


64 


STATICS. 


M         C 

N 

32 

» 

^^^\B 

<1 

\ 

f 

)V2' 

with  each  oth 
or  the  angle  a. 
Since  C  is  a 
fixed  point,  the 
only  condition 
of  equilibrium 
is,  that  the  sum 
of  the  moments 
about  C  is  zero  (Art.  75). 

Let  gz,  g2  be  the  centers  of  gravity  of  the  beams,  and  £jC 
=a,g2C=b.  Also,  w,=  weight  of  AC,  acting  at  g-^and  w2  = 
weight  of  BC,  acting  at  g2.  Draw  through  C  the  line  MCN 
horizontally,  meeting  the  vertical  directions  in  which  wx  and 
w2  act,  at  M  and  N. 

By(19),u)1CM-MJ3CN=0.  Let  BCN=0.  The  determin- 
ation of  6  will  fix  the  position  of  the  compound  beam. 

Since  CM=g1C.  cos.  MCA  and  CN=g-2C.  cos.  BCN,  we 
have 

w1.Cgl  cos.  ACM-w2.Cg2  cos.  BCN=0, 

0. 


or 


a.   cos.  (180  —  a+6)  —  w2.b.  cos.  6- 

_  wnb+w.a.  cos.  a 

.-.   Ian.  6=— - —. . 


wxa.  sin.  a 

Ex.  2.  When  a  given  weight  W  is  hung  from  the  end  of 

one  of  the  beams,  A  (Ex.  1),  find  6  in  case  of  equilibrium. 

ivn.b+(wi+2W)acos.  a 

Ian.  0= — 11—. ^r— - — ; . 

(w1+2W)asm.  a 

Ex.  3.  Two  beams,  as  in  Ex.  1,  are  suspended  from  one 

end  B ;  find  the  angle  0  which  the  upper  one  makes  with  a 

horizontal  line. 

(2ivx  +w2)b— wxa.  cos.  a 


Tan.  6=- 


w,a.  sin.  a 


N.  B.  Since  the  common  center 
of  gravity  of  the  two  beams  is  in  the 
vertical  through  B,  BM=Gm=Cm 
-CG=Cm-2NB=fl.  cos.  (a-0)- 
2b  cos.  0. 


EXAMPLES     ON     THE     PRECEDING     CHAPTERS.       65 


.*.  w2b.  cos.  6—wla.  cos.  (a  -6)+2wlb.  cos.  0=0. 

Ex.  4.  Two  spheres  of  unequal  radii,  but  of  the  same  ma- 
terial, are  placed  in  a  hemispherical  bowl ;  find  the  position 
they  take  when  in  equilibrium. 

Since  the  reactions  of  the  bowl  upon  the  spheres  are  in  the 
directions  of  the  radii  of  the 
spheres  through  the  points 
of  contact,  and  since  these 
radii  produced  pass  through 
C  ;  if  C  was  a  fixed  point, 
and  connected  with  A  and 
B  by  a  rigid  rod  without 
weight,  the  bowl  might  be 
removed  without  disturbing 
the  equilibrium.  The  ques- 
tion, then,  is  reduced  to  finding  the  position  of  equilibrium  ot 
two  weights  suspended  from  the  extremities  of  two  rigid  rods 
without  weight,  and  is  solved  like  the  preceding.  This  posi- 
tion will  be  known  when  0  is  known. 

Let  R  be  the  radius  of  the  bowl,  r,,  r2  the  radii  of  the 
spheres  A  and  B  respectively,  and  ACB=a. 

Then    AB=r1+r2>  CA=R-r1,  CB=R-r2, 
AC2+CB2-AB2 


M 

C 

1ST 

r 

Jo 

\    1       A 

/ 

B    )     / 

> 

1Vl 

Y 

-JV2 

and   cos.  a- 


2AC.BC       ' 
(R_ri)M-(R.-r,)"-<r1+r1>! 


,  which  gives  a. 


2(R-r1)(R-r2) 

Then  (19)  iu1.CM-u?3CN=0>  or  rJ.CM-»*CN=0 ;  since 
the  weights  of  the  spheres  are  as  the  cubes  of  their  radii. 
Substituting  the  values  of  CM=(R-r,)  cos.  (180— (a+0)) 
and  CN=(R  —  r2)  cos.  6,  expanding  and  reducing,  we  get 

Ian.  V= jy~ r— : . 

7j(lt— r,)  sin.  a 

Ex.  5.  A  heavy  beam  rests  upon  a  smooth  peg  with  one 
end  against  a  smooth,  vertical  wall ;  find  the  position  of  equi- 
librium. 

E 


66 


STATICS. 


Let  ACB  be  the  beam,  resting  at  A  against  the  wall  ADK 

and  upon  the  peg  C. 

The  center  of  gravity  g, 
when  there  is  an  equilibrium, 
will  evidently  be  at  some 
point  beyond  C  from  A.  Let 
Ag=a,  DC  =  &,  w=  the  weight 
of  the  beam  acting  at  g,  R= 
the  reaction  of  the  wail  per- 
pendicular to  itself  at  A,  ana 
R'=  the  reaction  of  the  peg 
perpendicular  to  the  beam  at 
C.  The  angle  6,  which  the  beam  makes  with  the  horizontal 
direction  when  in  equilibrium  by  the  action  of  these  three 
forces,  is  required. 

Employing  (17),  (18),  and  (19),  and  resolving  the  forces  in 
vertical  and  horizontal  directions,  and  about  the  point  C,  we 
have 

w    resolved  in  a  horizontal  direction  =     0, 
R  "  "  "  =     R, 

R'  "  "  "  =-R'sin.0. 

.-.  (17)  2.X  =  R-R'sin.  0=0.  (a) 

Also,  w  estimated  vertically  =     w, 

R  =     0, 

R'  "  "  =-R'cos.  0 

.-.  (18)  2.Y=w-R'cos.0  =  O.  (&) 

Also,  the  moment  pf  w  about  C=w.CF=w.(DF— DC). 
But  DF=AH  =  Ag-cos.0=acos.0,  and  DC =6. 
. .  w.CF  =  w.(a.  cos.  6— b), 
the  moment  of  R  about  C=R.CK=R.AK.tan.6»=R.&.  tan.0, 


R' 


o. 


(19)  2.(Xy-Y:c)  =  u>.(a.cos.0-&)— R.&.tan.0=O.    (c) 
Multiplying  (a)  by  cos.  0  and  (b)  by  sin.  0,  and  subtracting, 


we  have 


R.  cos.  0— w  sin.  0=0 
.-.  R  =  w  tan.  0. 


EXAMPLES     ON     THE     PRECEDING     CHAPTERS.      G'< 

Substituting  this  value  of  R  in  (c), 

w(a  cos.  6—b)  —  w.b.  tan.2  0=0, 

or  a  cos.  0=  ft  (1  +  tan.2  0)  =  ft.  sec.2  0=  b. — . 

v  '  cos."  0 


cos, 


3  lb 
0=  Vj  and  ft<a,  except  when  0=0 


.Ex.  6.  Solve  Ex.  5  by  resolving  the  forces  parallel  and  per- 
pendicular to  the  beam,  and  taking  the  moments  about  eithei 
A  or  g. 

The  three  forces  resolved  in  the-direction  of  the  beam  give 
R  cos.  6—w  sin.  0  =  0  ;  (a) 

resolved  perpendicularly  to  the  beam,  give 

R'-*ucos.0-Rsin.  0=0.  (b) 

The  moments  about  A  give 

R'.ft.  sec.  0— w.a.  cos.  0=0.  (c) 

From  (a),  we  have  R  =  w  tan.  0,  which,  substituted  in  (ftj, 
gives 

™  •     m        (cos.2  0+ sin.2  0) 

K'  =  uncos.  0+ tan.  0  sin.  6)=w j. -  =  w  sec.  0.    . 

cos.  0 

This  value  of  R',  substituted  in  (c),  gives 

.    ivb  sec.2  6—ioa  cos.  0  =  0. 


3  /ft 
V    a 


.*.  cos.  0=  \/-,  as  before. 
V    a 

The  moments  taken  about  g  give 

R'.Cg—  Rasin.  0  =  0, 
orR'(a  — ftsec.0)— R.asin.0— R'acos.0— R'ft  — Rasin.0cos.0— 0. 

Substituting  the  values  of  R  and  R'  above,  and  reducing, 

3  /ft 


cos.  0— 

v    a 

Ex.  7.  A  heavy  beam  lies  partly  in  a  smooth  hemispherical 
bowl  and  partly  over  the  edge  ;  find  the  position  of  equilibrium. 

The  beam  ABC  will  be  supported  by  the  reaction  R  of  the 
bowl  at  A,  perpendicular  to  the  surface,  or  in  the  direction  of 
the  radius  AO,  by  the  reaction  R'  of  ^?e  edge  of  the  bowl  al 


68 


STATICS. 


B  perpendicular  lo  tha 
beam,  and  by  its  weight 
w  acting  at  g. 

Let  Ag=a,  AO=r= 
radius  of  the  bowl,  and 
6 = ABO  =  BAO= incli- 
nation of  the  beam  to 
the  horizon. 

If  the  object  be  to 
determine  the  angle  6 
solely,  it  will  be  most 
readily  effected  by  resolving  the  forces  in  the  direction  of  the 
beam  and  taking  the  moments  about  B,  by  which  we  avoid 
expressions  involving  the  unknown  reaction  R'. 

The  reaction  R  in  the  direction  AB=     R  cos.  6, 
the  weight  w  "  "        =—ws'm.O. 

.'.  R  cos.  6— w  sin.  0=0, 
or  R  =  w  tan.  6. 

For  moments  about  B,  we  have 

R.AB  sin.  6—w.gB  cos.  0=0. 
But  AB=2r  cos.  6,  and  gB=(2r  cos.  6— a). 

.'.  2.R.r.  sin.  6  cos.  0— w>(2r  cos.  6— a)  cos.  6=0. 
Substituting  the  value  of  R,  and  reducing, 

2r  tan.  6.  sin.  6— 2r  cos.  0+a=O, 
2r—4r  cos.2  6+a  cos.  6=0. 


(a) 


(b) 


or 


cos.  6- 


a±V32ri+ai 

''  87 


in  which  the  +  sign  only  is  admissible. 

Ex.  8.  Solve  the  last  example  by  following,  step  by  step, 
the  method  of  Art.  71,  taking  A  for  the  origin  of  co-ordinates, 
and  AB  for  the  axis  of  x. 

Ex.  9.  Find  the  horizontal  strain  on  the  hinges  of  a  given 
door,  and  show  that  the  vertical  pressures  are  indeterminate. 

Let  the  annexed  figure  represent  the  door,  of  which  A  and 
B  are  the  hinges.  Let  g  be  its  center  of  gravity  at  which 
the  weight  w  acts.     The  door  is  kept  in  equilibrium  by  the 


EXAMPLES  ON  THE  PRECEDING  CHAPTERS.   O'J 


weight  acting  at  g,  and  the  reac- 
tions of  the  hinges  represented  by 
the  oblique  arrows  at  A  and  B. 

Let  A  be  the  origin  of  co-or- 
dinates, AX  the  axis  of  a:,  AY  the 
axis  of  y ;  and  let  x  =  a,  y=b  be 
the  co-ordinates  of  g ;  x  =  0,  y=h 
those  of  the  hinge  B.  Let  the 
resolved  parts  of  the  reactions  at 
B  beQ,,  horizontally,  and  Rl  vert- 
ically, and  Q2,  R2  those  at  A  re- 
spectively. 


Ri 


_S4- =a» 


A  Q2 


-X 


y, 


Then 


2.X  -Q.-Q^CorQ^Q,, 
2.  Y  =uj-R1—  Ra=0, 
S.(Xy-Ya;)=M).a-Q1A=0. 


(«) 
(6) 
(P) 


iv  a 


From  (a)  and  (c)  we  have  Q,=-t-  =  Q2' 

which  gives  the  horizontal  strain ;  and  it  is  the  same  in  mag 
nitude  at  each  hinge,  but  opposite  in  direction. 

Again,  from  (b)  we  have  R,+R2=i«,  but  wTe  have  no  oth- 
er relation  by  which  we  may  determine  the  values  of  Rj  and 
R2,  which  are  therefore  indeterminate. 

Ex.  10.  Two  given 
smooth  spheres  rest 
in  contact  on  two 
smooth  planes,  inclin- 
ed at  given  angles  to 
the  horizon  ;  find  their  2 
position  of  equilibri- 
um. 

Let  the  planes  AB,  AC  make  the  angles  a  and  0  respective- 
ly with  the  horizontal  line  through  A;  O,  and  02  be  the  cen 
ters  of  the  spheres  at  which  their  weights  wy,w2  respective!) 
act:  Rj,  R9  the  reactions  of  the  planes  at  the  points  of  con 


70 


S  T  A  T  1  t:  s. 


tact,  perpendicular  to  themselves,  and  therefore  passing  through 
the  centers  of  the  spheres  to  which  they  are  tangents.  Let  S 
be  equal  to  the  mutual  pressure  of  the  spheres  at  their  point 
of  contact,  acting  in  the  line  passing  through  their  centers,  and 
making  the  angle  6  with  the  horizontal  line  AD.0  is  required. 
Each  sphere  is  in  equilibrium  from  its  own  weight,  the  re 
action  of  the  plane  against  which  it  rests,  and  the  pressure  of 
the  other  sphere. 

By  resolving  the  forces  in  the  direction  of  each  plane  for  the 
equilibrium  of  each  sphere,  we  shall  avoid  equations  involving 
the  unknown  reactions  R,  and  R„,  and  have,  in  the  direction 
of  AB, 

w2  sin.  a— S  cos.  (a—6)=Q,  (a) 

in  the  direction  of  AC, 

w1  sin.  0-S  cos.  (0+0)  =  O.  (b) 

Eliminating  S,  we  have 

w2  sin.  a.  cos.  (j3+0)  =  i«1  sin.  (3.  cos.  (a— 6). 
Expanding  and  reducing, 

w0  sin.  a.  cos.  (3.  — w.  sin.  /3.  cos.  a    w0  cot.  jQ— wl  cot.  a 

tan.  6  =  — 7 ; r— : : 5 =— : . 

(w1JrW2)  sin.  a.  sin.  p  iv1+w2 

Ex.  11.  A  sphere  is  sustained  upon  an  inclined  plane  by  the 
pressure  of  a  beam  movable  about  the  lowest  point  of  the  in- 
clined plane ;  given  the  inclination  of  the  beam  to  the  plane, 
required  that  of  the  plane  to  the  horizon. 

Let  AgB  be  the 
beam  movable  about 
A,  w=  weight  of 
beam  acting  at  g,  B 
-Pj  the  point  of  contact 
with  the  sphere 
whose  center  is  C, 
m>'==  weight  of  the 
sphere. 

The  sphere  is  in  equilibrium,  from  the  reaction  R  of  the 
plane  at  the  point  of  contact  F,  from  the  pressure  P  of  the  beam 
at  B,  and  from  its  own  weight  w'.  These  three  forces  all  act 
through  the  center  C.     Ag=a,  AB  =  b,  BAD  — a  are  given,  or, 


EXAMPLES     ON     THE     PRECET/NG     CHAPTERS.     71 


instead  of  either  of  the  two  latter,  the  radius  of  the  sphere  may 
be  given.  The  angle  DAE  =  0,  the  inclination  of  the  plane 
when  in  equilibrium,  is  required.  » 

For  the  condition  of  equilibrium  of  the  beam,  take  the  mo- 
ments about  A. 

"P.AB=w.Ag.  cos.  (a+6). 

Hence  ^~~h  cos"  (a~^)* 

For  the  condition  of  equilibrium  of  the  sphere,  resolving  the 
forces  in  the  direction  of  AD,  we  have 

w'  sin.  6— P  sin.  a=0, 


or 


whence 


u   • 
w'  sin.  6—Wj  sin.  a  cos.  {a-\-0)  ; 


tan.  6- 


wa  cos.  a.  sin.  a 


wa  sin.2a+u;'6' 

which  gives  6  the  elevation  of  the  plane,  as  required. 

Many  statical  problems  require,  for  the  determination  of  all 
the  unknown  quantities,  equations  to  be  formed  by  geometrical 
relations.     Take,  for  illustration,  the  following : 

Ex.  12.  A  heavy  beam  turns  about  a  hinge,  and  is  kept  in 
equilibrium  by  a  cord  attached  to  the  lower  end  ;  the  cord 
passes  over  a  pulley  in  the  same  horizontal  line  with  the  hinge, 
and  sustains  a  given  weight ;  find  the  position  of  equilibrium 
of  the  beam. 

Let  A  be  the  hinge,  C  the 
pulley,  AC=c,  AB  the  beam 
=1  in  length,  g  its  center  of 
gravity  at  which  its  weight 
w  acts,  Ag=a  and  P=  the 
weight  hung  from  the  cord 
and  measuring  its  tension  t. 

Let  6>=ABC  and  £=BAC,    ® 
both  unknown. 

Taking  moments  about  A,  we  have, 

t.AB.  sin.  6=w.Ag  cos.  (t>, 


72 


STATICS. 


or 


w  a 
tin.  0=^.-.  cos.  0. 


PT 

From  the  geometrical  data,  we  have 

AC_    sin.  0 
AB~sin.ACB' 
c 


(a) 


or 


sin.  0=7-  sin.  (0+0). 


(*; 


Equations  (a)  and  (b)  suffice  to  determine  0  and  0. 
i?x.  13.  A  uniform  beam  rests  with  its  lower  end  in  a 
smooth,  hemispherical   bowl,  and   its    upper   end   against  a 
smooth,  vertical  plane.     Find  the  position  of  equilibrium. 

The  beam  AB  rests  against  the 
^  vertical  plane  AD  at  A  and  upon 
the  bowl  at  B,  and  is  sustained  by 
its  weight  w  acting  at  g,  the  reac- 
}  tion  R  of  the  bowl  in  the  radius 
BC,  and  the  reaction  R'  of  the 
plane  perpendicular  to  itself. 

Let  r=  radius  of  the  bowl ;  AB 
=2a,  Ag=a,  since   the    beam   is 
uniform,  w=  weight,  and  CD=d, 
ad  supposed  known. 

Resolving  vertically,  we  have 


R  sin.  O—w—Q,  or  R: 


or 


sin.  0' 
Taking  the  moments  about  A,  we  have 

R.AB.  sin.  (0— 0)  — w.Ag-.  cos.  0=0, 
R.2a.  sin.  (0—<p)  —  w.a.  cos.  0=0. 
sin.  (0—0)     cos.  0 


(a) 


sin.  0  2 

or  cos.  <ff—  cot.  0.  sin.  0— ^  cos.  0=0. 

Hence  tan.  0=2  tan.  0. 

This  equation  containing  two  unknown  quantities,  a  geo- 
metrical relation  between  them  must  be  obtained.  Cm  being 
«  vertical  line  meeting  AB  in  m, 


EXAMPLES     ON     THE     PRECEDING     CHAPTER 


73 


cos.  0 

COS.  (p 


Bm 
BC: 


2a— Am     2a— CD.  sec.  0    2a— dsec.  0 


cos.  6- 


2a  cos.  0— d 


ih) 


Equations  (a)  and  (b)  are  sLfficient  to  determine  0  and  0,  aa 
required. 

Ex.  14.  A  heavy  beam  has  one  end  resting  against  a  smooth 
vertical  wall,  and  the  other  sustained  by  a  cord,  which  is  fast- 
ened at  a  point  vertically  above  the  point  where  the  beam 
rests.    Find  all  the  forces  which  keep  the  beam  in  equilibrium. 

Let  CB  be  the  beam,  AB  the  cord,  A     ^ 
and  C  the  points  on  the  vertical  wall  AD. 

Let  w—  weight  of  the  beam,g-  its  center 
of  gravity  Cg—a,  CB=/,  AB=c,  and  AC 
=h,  all  supposed  known.  The  angles  A, 
B,  and  C  will  be  known. 

Let  t—  tension  of  the  cord.  The  beam 
will  press  against  the  wall,  and  this  press- 
ure may  be  resolved  in  a  vertical  and  hor- 
izontal direction  ;  the  latter,  perpendicular 
to  the  wall,  will  be  destroyed  by  its  reac- 
tion ;  but,  since  the  wall  is  smooth,  the 
vertical  component  can  be  balanced  only 
by  an  opposite  force  =P.  This  vertical 
omponent  will  be  upward,  or  downward,  or  zero,  according 
to  the  position  of  the  point  C.  We  shall  suppose  its  position 
to  be  such  that  the  component  may  be  upward,  and  require  a 
force  to  be  applied  downward  to  keep  the  end  of  the  beam 
atC. 

Resolve  vertically  and  horizontally,  and  take  the  moments 
about  C. 

P+/  cos.  A  —  iv=0,  (a) 

R-ts'm.  A  =  0,  (b) 

w.a.  sin.  C  —  t.L  sin.  B=0.  (c) 

w.a.  sin.  C     w.a.c 

tension  of  cord. 


From  (c), 


t=- 


w.a.c 
I.  sin.  B  l.h 

Substituting  the  value  of  /  in  (b), 


74 


STATICS. 


CLC 

R=Wjr.  sin.  A=  pressure  aga  nst  the  wall ; 


IK 

and  from  (a), 

V=w  (  1  — —  cos.  A  )  =  force  to  be  applied  at  C, 

to  prevent  the  beam  from  sliding  along  the  wall. 

Ex.  15.  A  weight  w  hangs  from  one  end  of  a  cord  of  which 
the  other  end  is  fastened  to  a  vertical  wall ;  the  cord  is  pushed 
from  the  wall  by  a  rod  tied  to  it,  which  is  perpendicular  to  .the 
wall.  Find  the  pressure  R  of  the  rod  on  the  wall  when  the 
cord  makes  the  angle  a  with  the  wall. 

Ans.  R=u>.  tan.  a. 

Ex.  16.  A  heavy  beam,  AB  =  Z,  of  which  the  weight  is  w, 
lies  with  the  end  A  against  a  smooth  vertical  wall  AD,  and 
the  end  B  on  a  smooth  horizontal  plane  DB,  making  with  it 
the  angle  6.  The  distance  of  its  center  of  gravity  g  from  B  is 
a,  and  it  is  kept  in  equilibrium  by  a  cord  attached  to  it  at  B 
and  fastened  at  D.  Required  the  tension  of  the  cord,  the  re- 
action R  of  the  wall,  and  the  pressure  P  on  the  plane. 

Ans.  R  =  t  =  w.j.  cot.  6.     V=w. 

Ex.  17.  A  body  (weight  =w)  is  suspended  by  a  cord  (length 

=  /)  from  the  point  A  in  a  horizontal  plane,  and  is  thrust  out 

of  its  vertical  position  by  a  rod  without  weight,  acting  from 

another  point  B  in  the  horizontal  plane,  such  that  AB  =  d,  and 

making  the  angle  6  with  the  plane.     Find  the  tension  t  of  the 

cord. 

I 
t=w- cot.  6. 
a 

Ex.  18.  A  triangular  plate  of  uniform  thickness  and  density 
is  supported  horizontally  by  a  prop  at  each  angle.  Find  the 
pressure  on  each  prop. 

Ex.  19.  A  uniform  beam  rests  on  two  planes  inclined  at  an- 
gles a  and  fi  to  the  horizon.  Find  the  inclination  0  of  the  beam 
to  the  horizon. 

sin.  (0—a) 
tan.0=      .  . 

2  sin.  a  sin.  p 


£ 


%cl  >»  z^6c  / 7 


-3 


/—    // 

.    '.    £~=-  J?   e  w  e  ( / ) 


. 


5L2 


EXAMPLES     ON     THE     PRECEDING     CHAPTERS. 


Ex.  20.  A  uniform  beam  AB  hangs  by  a  string  BC  from  a 
fixed  point  C,  with  its  lower  extremity  A  resting  on  a  smooth, 
horizontal  plane.  Show  that,  when  there  is  an  equilibrium,  CB 
must  be  vertical. 

Ex.  21:  A  uniform  beam  AB  is  placed  with  one  end  A  in- 
side a  smooth  hemispherical  bowl,  with  a  point  P  resting  on 
the  edge  of  the  bowl.     If  AB=  3  times  the  radius  R,  find  AP. 

AP=1.838R. 

Ex.  22.  A  beam,  whose  weight  is  w  and  length  6  feet,  rests 
on  a  vertical  prop  CD  (=  3  feet) ;  the  lower  end  A  is  on  a  hor- 
izontal plane,  and  is  prevented  from  sliding  by  a  string  DA 
(=  4  feet).     Find  the  tension  of  the  string. 

lension  =— — w. 
125 

Ex.  23.  A  uniform  beam,  CB=2a,  has  one  end  C  resting 
against  a  smooth,  vertical  wall  AC,  and  the  other  sustained  by 
a  cord,  whose  length  is  c,  fastened  to  the  point  A.  Find  AC 
when  the  beam  is  in  equilibrium  from  its  weight  w,  the  tension 
t  of  the  cord,  and  the  reaction  R  of  the  wall.     (Vide  Ex.  14.) 

Ex.  24.  Given  the  inclination  i  of  the  right  line  AC  to  the 
horizon  and  the  weight  of  a  heavy  body  W,  to  determine  what 
force  or  weight  Pj  acting  in  the  given  direction  WM  will  be 
sufficient  to  sustain  W  on  the  line. 

The  body  W  is  kept  upon  the  line  by  the  action  of  the  force 
Pj  in  the  direction 
WM,  and  its  weight 
acting  in  the  direction 
WE.  The  reaction  of 
the  line  normal  to  its 
direction  is  a  third 
force  (Art.  76). 

Resolving  P  and  W 
horizontally  and  vert- 
ically, we  have 


E|    D 


S.X=P,  cos.  (e+i)+Wcos.  270°=P,  cos.  (e+i), 
S.Y=P1  sin.  (e  +  i) +W cos.  180°  =  P1  sin.  (e+i)- W 


76  STATICS. 

Hence  (20)  becomes 

P,  cos.  (e+z*)  +  (P1  sin.  (e+i)-W)  tan.  i=0; 

or,  expanding  and  reducing, 

P,  cos.  e=W  sin.  i, 

Ws'm.i 
or  P,= .  (a) 

cos.  e 

If  the  reaction  of  the  line  be  required,  resolve  all  the  forces 
horizontally  (8).     This  gives 

N.  cos.  (90°  +  i)+P1  cos.  (e+i)  =  0. 
^  cos^+0 

1      sin.  i  v  ' 

or,  by  (a),  N=W.^hSS+l>.  (e) 

J  v  '  cos.  e  v  ' 

Ex.  25.  A  given  weight  W  is  kept  at  rest  on  a  circular  arc 
by  a  weight  P  attached  to  a  cord  which  pass- 
es over  a   point  M  in  the  vertical  line  MX 
iw  through  C,  the  center  of  the  circle.     Required 

the  position  of  W,  supposing  no  friction  at  M. 

Resolving  the  forces  P  and  W  vertically  and 
horizontally,  we  have,  calling  the  angle  PMW,  e, 
2.X=W-Pcos.e, 
2.Y=      -P  sin.  e. 
Let  the  co-ordinates  of  the  point  W  be  x,  y,  and  call  the  dis- 
tance MC,  a,  and  MW,  I ;  then  the  equation  referred  to  M  as 
the  origin,  is 

(a-xy+y*=r\ 
or  y—^/ri  —  {a—x)n- 

Differentiating,  we  obtain 

dy  a—x  a—x 


dx      Vr2-(a-.r)2        V 

x 

Substituting  in  (21),  and  recollecting  that  cos.  «=7  and 


sin.  e=j,  we  get 


w_P^_pL^=w._P:L_p^=o 

l         Idx  I  I 


Hence 


EXAMPLES     ON     THE     PRECEDING     CHAPTERS.      77 

aP 


b 


W 


an  equation  which  determines  the  distance  of  W  from  M. 

M 
Ex.  26.  Instead  of  a  circle,  as  in  Ex.  25,  let  the 
curve  be  a  hyperbola  with  its  transverse  diameter  m 
vertical,  the  point  M  being  at  its  center. 


Ex.  27.  Required  a  curve  such  that  a  given 
weight  P,  by  a  cord  passing  over  a  fixed  point   m 
without  friction,  will  balance  another  given  weight 
W  at  every  point  of  it. 


W 


Ex.  28.  Required  the  co-ordinates  of  the  center  of  gravity 
of  a  semiparabola  whose  equation  is  y°=px,  height  =a,  and 
base  =b. 

Ans.  x=%a,  y—lb. 

Ex,  29.  Required  the  center  of  gravity  of  the  surface  of  a 
right  cone. 

Ex.  30.  Find  the  center  of  gravity  of  a  paraboloid  of  revo- 
lution whose  altitude  is  a. 

Ans.  x=fa. 

Ex.  31.  Find  the  center  of  gravity  of  a  segment  of  a  hyper* 
boloid  whose  altitude  is  a. 


Jr 


j> 


£ 


W 


CHAPTER   VI. 

ON     THE     MECHANICAL     POWERS. 

130.  The  general  object  of  machinery  is  to  transmit  and  to 
economize  the  action  of  certain  forces  at  our  disposal.  The 
specific  end  is,  sometimes  to  augment  the  action  of  which  the 
power  employed  is  capable  when  applied  without  the  inter- 
vention of  machinery  ;  sometimes,  merely  to  change  the  direc- 
tion of  the  action ;  and  sometimes  to  regulate  the  velocity  of 
the  point  to  which  the  action  is  transmitted. 

The  most  simple  machines  are  denominated  Mechanical pow 
ers,  and  are  reducible  to  three  classes,  viz.,  the  Lever,  Cord. 
and  Inclined  Plane. 

The  first  class  comprehends  every  machine  consisting  of  a 
solid  body  capable  of  revolving  on  an  axis,  as  the  Wheel  and 
Axle. 

The  second  class  comprehends  every  machine  in  which 
force  is  transmitted  by  means  of  flexible  threads,  ropes,  &c, 
and  hence  includes  the  Pulley. 

The  third  class  comprehends  every  machine  in  which  u 
hard  surface  inclined  to  the  horizon  is  introduced,  as  the  Wedge 
and  the  Screw. 

The  force  which  is  used  to  sustain  or  overcome  any  oppo- 
sition is  called  the  Power ;  the  opposition  to  be  overcome  is 
called  the  Weight.  This  distinction  in  the  names  of  the  forces 
employed  implies  none  in  their  nature. 

§  I.    THE  LEVER. 

131.  Def.  A  Lever  is  an  inflexible  rod  capable  of  motion 
about  a  fixed  point,  called  a  fulcrum.  The  rod  may  be  straight 
or  any  how  bent. 

It  is  generally  regarded,  at  first,  as  without  weight,  but  its 


THE     LEVER. 


79 


weight  may  obviously  be  considered  as  another  force  applied 
in  a  vertical  direction  at  its  center  of  gravity. 

Def.  The  arms  of  a  lever  are  the  portions  of  it  intercepted 
between  the  power  and  fulcrum,  and  between  the  weight  and 
fulcrum. 

132.  Levers  are  divided  into  three  kinds,  according  to  the 
relative  positions  of  the  power,  weight,  and  fulcrum. 

In  a  lever  of  the 
first  kind,  the  fulcrum 
lies  between  the  points 
at  which  the  power 
and  weight  act. 


In  a  lever  of  the  third 
kind,  the  point  of  action  of 

the  power  is  between  that      B ] 2 

of  the  weight  and  the  ful-  -A- 

crum. 


In  a  lever  of  the  sec- 
ond kind,  the  weight  acts 
at  a  point  between  the 
fulcrum  and  the  point  of 
action  of  the  power. 


133.  Prop.  Required  the 
condition  of  equilibrium 
and  pressure  on  the  ful- 
crum when  two  parallel 
forces  act  on  a  straight 
lever. 


A^ 


HO 


STATICS. 


Since  the  fulcrum  C  is  a  fixed  point,  by  Art.  75,  the  sum  ol 
the  moments  of  the  forces  about  C  must  be  zero. 

Aw  Let  a  be  the  angle  made  by 

/  the.  direction  of  the  forces  with 

A.  /  V7  the    lever.      From    (19),    we 

B  C    have 

W.BC.  sin.  a— P.AC.  sin.  a=o, 

the  moment  of  P  being  nega- 
rP  tive,  since  it  tends  to  produce 

motion  in  a  direction  opposite  to  that  of  W. 

W    AC 

Hence  -5-=^,  or,  in  case  of  equilibrium,  the  weight  and 

power  are  reciprocally  proportional  to  the  distances  at  which 
they  act  from  the  fulcrum. 

Hence  ( Art.  36)  the  resultant  of  P  and  W  must  pass  through 
C,  and  the  pressure  on  the  fulcrum  is  equal  to  the  algebraic 
sum  of  P  and  W,  and  acts  in  the  direction  of  the  greater. 

Cor.  If  the  power  equal  the  weight,  the  distances  of  their 
points  of  action  from  the  fulcrum  will  be  equal. 

For  P=W  gives  AC=BC. 

134.  Prop.  Required  the  condition  of  equilibrium  and  the 
pressure  on  the  fulcrum  when  any  two  forces  in  the  same  plane 
act  on  a  straight  lever. 

jy  Let   the   forces   P 

^  "'     /  \\  and  W  make  the  an- 

/'       \  \  gles  a  and  /3  respect- 

/             '•   \  ively  with  the  lever, 

/                  \     \  and   let  their   direc- 

/\                  q/\  -X  tions,  when  produced 

a  /     ':                     C  \    i    \-q  if  necessary,  meet  in 


^T         \  D.    Since  C  is  a  fixed 


1  w    point  in  the  lever,  the 

*  R'  sum  of  the  moments 

wf  P  and  W  about  C  must  be  zero.     Hence 

W.CB.  sin.  /3-P.AC.  sin.  a=0, 


THE     LEVER. 


W    AC.  sin.  a 


or 


(2.) 


\a 


P      CB.  sin.  ft' 

But  AC.  sin.  a=  per- 
pendicular from  C  on 
AD,  and  CB.  sin.  0= 
perpendicular  from  C 
onBD.  Hence  the  con- 
dition of  equilibrium 
requires  that  the  pow-  j±  ,-''' 

er  and  weight  should  be 
inversely  as  the  perpen- 
diculars from  the  ful- 
crum on  their  respective  directions. 

Since  the  lever  is  in  equilibrium  by  the  actions  of  P  and  W 
and  the  reaction  of  the  fulcrum,  the  resultant  of  P  and  W  must 
be  equal  and  opposite  to  that  reaction.  It  will  therefore  pass 
through  C,  and  be  equal  to  the  pressure  on  the  fulcrum. 

To  find  R,  we  have  {Art.  29) 

R2=P2+W2+2PW.  cos.  ADB   in  Fig.  (1), 
and         R2=P2+W2+2PW  cos.  ADW  in  Pig.  (2). 

But    ADB=180-(a+j3),    and    ADW=180-ADB=180- 
(P-a). 

...  R2=P+W2-2PWcos.(a+0), 
or  R2=P2+W2-2PW  cos.  (0-a). 

To  find  the  angle  6  made  by  R  with  the  lever,  resolve  par- 
allel and  perpendicular  to  the  lever,  the  reaction  of  the  fulcrum 
being  equal  and  opposite  to  R.     Hence  we  have,  in  Fig.  (1), 

P.  cos.  a— W  cos.  j3+R  cos.  0=0, 
P   sin.  a+W  sin.  /3-R  sin.  0=0. 


Tan.  6= 


P.  sin.  a+W  sin.  (3 


in  Fig.  (2), 


W  cos.  j3— P  cos.  a  ' 

P  cos.  a— W  cos.  /3— R  cos.  6=0, 
P  sin.  a-  W  sin.  /3+R  sin.  6=0. 

W  sin.  (3— P  sin.  a 


Tan.  6= 


P  cos.  a—W  cos.  (3' 
F 


82 


S  T  A  1  ICS. 


135.  Cor.  Whenever  the  lever  is  bent  or  curved,  the  conch 
tion  of  equilibrium  is  the  same. 

For,  since  the 
moments  of  the 
forces  about  the 
fulcrum  in  oppo- 
site directions  must 
be  equal,  in  case 
of  equilibrium,  we 


have 


or 


P.CM=W.CN. 
WCM 
T~CN' 


136.  Prop.  Required  the  condition  of  equilibrium  and  the 
pressure  on  the  fulcrum,  when  any  number  of  forces  act  in  any 
direction  in  one  plane  on  a  lever  of  any  form. 

Let  Pj,  P2,  P3,  &c,  be  the  forces  in  one  plane, 

Pi,p2,p3,     "     the  distances  of  their  points  of  applica 

tion  from  the  fulcrum, 
a,,  a„,  a3,     "     the  angles  made  by  the  directions  of  P 
and  p  respectively. 

Then  px  sin.  a1,p2  sin.  a2,  p3  sin.  a3  are  the  perpendiculars 
from  the  fulcrum  on  the  directions  of  the  forces,  and  Pj p^  sin. 
a,,  &c,  the  moments  of  the  forces. 

When  there  is  an  equilibrium,  the  sum  of  the  moments  of 
the  forces  about  the  fulcrum  will  be  zero,  or 

Vlpl  sin.  «i+P2p2  sin.  a2+F3p3  sin.  a3  +  ,&c.,  =0. 

The  signs  of  the  moments  will  depend  on  the  direction  in 
which  they  tend  to  produce  rotation. 

To  find  the  pressure  on  the  fulcrum,  we  must  determine  the 
resultant  of  all  the  forces  supposed  to  concur  at  the  fulcrum. 
For  each  moment  F1p1  sin.«,  is  equal  to  a  couple  of  which 
the  forces  are  P  and  the  arm  p  sin.  a,  and  a  single  force  P  act 
ing  at  the  fulcrum.  The  magnitude  of  R  will  then  be  determ- 
ined by  the  equation  (0), 

R=v/(2.X)2  +  (2.Y)2. 


THE     LEVER.  83. 

Cor.  If  the  lever  be  straight  and  the  forces  parallel, 

a1=a2=a3  =  ,  &c. 
Hence  P1je>1+P2^2+P3^3+,  &c.,=0. 

The  same  result  will  also  be  obtained  by  following,  step  b.v 
step,  the  method  of  Art.  71. 

EXAMPLES. 

Ex.  1.  On  a  straight  lever  AB  of  the  first  kind,  without 
weight,  36  inches  in  length,  a  weight  W=15  lbs.,  acting  at  B, 
is  balanced  by  a  power  P=3  lbs.  acting  at  A.  Required  the 
distance  of  the  fulcrum  C  from  A. 

LetAC=x;  then  BC  =  3G-x. 

By  Art.  133,         I\AC=W.BC, 
or  3.z=15(36-x). 

Hence  .t=30  inches  =AC. 

Ex.  2.  On  one  arm  (=p1)  of  a  straight  lever  of  the  first 
kind,  without  weight,  a  body  counterpoises  a  weight  (=a  lbs.), 
on  the  other  (=p2)  a  weight  (=b  lbs.).  Required  the  weight 
of  the  body. 

A  straight  lever  of  the  first  kind,  with  unequal  arms,  and  hav 
rag  the  fulcrum  at  its  center  of  gravity,  is  called  a  false  balance. 

Let  x—  the  unknown  weight. 

By  Art.  133,  x.p^—a.p. 

and  x.p2=b.p 

By  multiplying  the  equations  member  by  member, 

x*=ab, 
x  =  Vab. 
Ex.  3.  On  a  straight  lever,  without  weight,  are  suspended 
five  bodies,  Pj-4  lbs.,  P2  =  10  lbs.,  P3  =  2  lbs.,  P4=3  lbs.,  P5 
=7  lbs.,  at  the  points  A,  B,  C,  D,  and  E,  such  that  AB=4  feet, 
BC=2  feet,  CD=6  feet,  and  DE=8  feet.  Required  the  posi- 
tion of  the  fulcrum  F,  about  which  they  will  balance. 

A B         C  D E 


'2» 


84  STATICS. 

Let  AF— x.     Then,  by  Art.  136,  Cor.,  we  have 
P,=  4and^,=AF=.r.  .\Vlpl  =  4.x. 

P2  =  10  «  jo„=BF=AF-AB=x-4.  P2jpa=10(z-  4). 
p3=  2  "  /3=CF=AF-AC=:c-6.  P3/>3=  2(x-  G) 
P4=  3  "  p4=DF=AF-AD=.x-12.  P4jp4=  3(x-I2) 
P.=  7    "   j95=EF=AF-AE=x-20.       P5p5  =  7(z-20). 

And  2.Pp=4x  +  10(x-4)+2(a;-6)  +  3(.r-12)+7(x-20)=0 
=26:c— 228. 
Hence  z=AF=8if  feet. 

Otherwise :  Since  the  weights  are  parallel  forces,  their  re- 
sultant R  is  equal  to  their  sum.  The  whole  system  being  in 
equilibrium,  the  resultant  must  pass  through  the  fulcrum,  and 
the  moment  of  the  resultant  must  be  equal  {Art.  45)  to  the  sum 
of  the  moments  of  the  components.  Taking  A  for  the  origin 
of  moments,  we  have 

R.x=P1.0+P2.AB+P3.AC+P4.AD+P5.AE, 

or    26.^=4X0+  1X4  +  2X6  +3X12+7X20=228. 
.-.  z=AF=8ff  feet. 

Ex.  4.  A  uniform  lever  AB  of  the  first  kind,  12  feet  long, 
whose  weight  w—Q  lbs.,  has  a  weight  W=100  lbs.  suspended 
from  the  shorter  arm  CB=2  feet.  Required  the  power  P 
which  must  be  applied  vertically  at  A,  to  equilibrate  W. 

The  weight  of  the  lever  has  the  effect  of  a  weight  equal  to 
itself,  applied  at  its  center  of  gravity,  which,  since  the  lever  is 
uniform  in  size  and  density,  is  at  its  middle  point.     The  dis- 

Ap "Dp 

tance  of  this  point  from  C  is ,  and  its  moment  about 

r      nu       AC~BC 
G  will  be  w. . 

Hence  P.AC+w- ~— =W.BC, 

W.BC->C-BC)     100X2_3X8 
or        P= m = =  17.6  lbs. 

Ex.  5.  The  arms  of  a  bent  lever  are  a=3  feet  and  b=5  feet, 
and  inclined  to  each  other  at  an  angle  0=150°.     To  the  arm 


THE     LEVER. 


85 


a  a  power  P=7  lbs.  is-  applied,  and  to  the  arm  b  a  weight  W 
=  6  lbs.  Required  the  inclination  of  each  arm  to  the  horizor 
when  there  is  an  equilibrium. 


Let  a  be  the  inclination  of 
the  arm  a,  and  (3  the  inclina- 
tion of  the  arm  b  to  the  ho- 
rizon. 


Then  P.MC  =  W.NC, 
or  P.a.  cos.  a=W.b.  cos.  p=W.b.  cos.  (180°- {a+6)), 
=  -W.b.cos.(a+6)  ' 
=  —  W.b.  cos.  a  cos.  0+W.&.  sin.  a.  sin.  0. 
P.a+W.&.cos.0     21+30.  cos.  150° 


wQ 


Tan.  a- 


W.b.  sin.  6 
21-30  cos.  30° 


30.  sin.  150° 
21-30xiv/3 


4,98 
~15 


30.  sin.  30  30Xf 

Log.  Tan.  «=9.5211381n, 

a=-18°.22', 
and  0=180°- (a+6»)  =  180o-r-18o.22'-150o=48°.22 . 

Hence  the  arm  AC  is  inclined  at  an  angle  of  18°.22'  above 
the  horizon,  and  BC  at  an  angle  of  48°.22'  below  the  horizon. 
Ex.  6.  The  whole  length  of  the  beam  of  a  false  balance 
(Ex.  2)  is  3  feet  9  inches.  A  body  placed  in  one  scale  coun- 
terpoises a  weight  of  9  lbs.,  and  in  the  other  a  weight  of  4  lbs. 
Required  the  true  weight  W  of  the  body,  and  the  lengths  a  and 
b  of  the  shorter  and  longer  arms. 

Ans.  W=G  lbs.,  a=\  ft.  G  in.,  6=2  ft.  3  in. 
Ex.  7.  A  false  balance  has  one  of  its  arms  exceeding  the 

1  • 

other  by  — th  part  of  the  shorter  arm.  Supposing  a  shop- 
keeper, in  using  it,  puts  the  weight  as  often  in  one  scale  as  the 
other,  does  he  gain  or  lose,  and  how  much  per  cent.  1 

Ans.  Loses  — r- —  per  cent. 
m  +m 


V-w;_  ( 


\A/~   Vv_ 


XU^JL        ^-&L 


JlS 


S6 


STATICS. 


Ex.  8.  The  anus  of  a  bent  lever  are  3  feet  and  5  feet,  and 
inclined  to  each  other  at  an  angle  of  150°,  and  to  the  shorter 
arm  is  suspended  a  weight  of  7  lbs.  Find  what  the  othei 
weight  must  be  in  order,  1st,  that  the  shorter,  and  2d,  that  the 
longer  arm  may  rest  in  a  horizontal  position. 

Ans.   1st,  4.85  lbs. 
2d,  3.64  lbs. 


§  II.    WHEEL    AND    AXLE. 

137.  The  Wheel  and  Axle  consists  of  a 
cylinder  or  axle,  perpendicular  to  which  is 
firmly  fixed  a  circle  or  wheel,  whose  center 
is  in  the  axis  of  the  cylinder.  The  whole  is 
supposed  to  be  perfectly  rigid,  and  movable 
only  round  the  axis  of  the  cylinder. 

In  the  ordinary  applications  of  this  ma- 
chine, the  power  is  applied  tangentially  to 

the  surface  of  the  wheel,  and  the  weight,  in  the  same  manner, 

to  the  surface  of  the  axle. 

138.  Prop.  Required  the  condition  of  equilibrium  of  the  wheel 
and  axle  when  two  forces  are  applied  tangentially  to  the  circum- 
ference of  the  axle  and  of  the  wheel. 

Let  ADE  be  a  section  of  the 
wheel  perpendicular  to  the  axis, 
and  BCA  a  horizontal  line 
through  the  center  of  the  axis, 
terminated  at  A  by  the  circum- 
ference of  the  wheel,  and  at  B 
by  the  circumference  of  the 
axle.  The  power  P  acts  in  this 
vertical  plane  at  A,  and  the 
weight  W  acts  in  a  plane  par- 
allel to  it.  Since  the  axle  and 
wheel  are  firmly  connected,  the 
action  of  the  weight  may  be 
transferred  to  the  plane  ADE 


WHEEL     ArfD     AX  I.E.  81 

{Art.  57),  and  supposed  to  act  at  B.     Then,  since  the  sum  of 
the  moments  must  be  zero  for  equilibrium, 

P.AC-W.BC=0. 

W_AC__radius  of  wheel 
P     BC      radius  of  axle   ' 

or,  the  Power  :  Weight— radius  of  axle  :  radius  of  wheel. 

139.  Cor.  1.  The  same  relation  will  exist  in  all  positions  ot 
the  wheel,  so  that  this  machine  may  be  called  a  perpetual  lever. 

140.  Cor.  2.  When  the  power  and  weight  act  vertically  on 
opposite  sides  of  the  axis,  the  pressure  on  the  rests  or  Ys 
equals  the  sum  of  the  two ;  when  vertically  on  the  same  side, 
it  equals  their  difference ;  when  in  any  other  directions,  it 
equals  the  diagonal  of  a  parallelogram,  whose  sides  represent 
the  power  and  weight  in  magnitude  and  direction. 

If  ropes  are  used  to  transmit  the  action  of  the  power  and 
weight,  we  must  suppose  the  forces  applied  to  the  axes  of  the 
ropes.  Hence,  if  r  and  R  represent  the  radii  of  the  axle  and 
wheel  respectively,  and  t  and  T  represent  half  the  thickness 
of  the  ropes. 

W.(r+0=P-(R+T), 
or  P  :  W=r+t  :  R  +  T. 

141.  Prop.  Required  the  condition  of  equilibrium  of  any  num- 
ber of  forces,  acting  in  any  direction  in  planes  perpendicular  to 
the  axis. 

Let  the  actions  of  all  the  forces  be  transferred  (Art.  57)  to 
the  same  plane ;  then,  since  the  intersection  of  the  axis  with 
this  plane  is  a  fixed  point,  the  condition  of  equilibrium  is  given 
by  Art.  74  ;  or,  the  algebraic  sum  of  the  moments  of  all  the  forces 
about  the  axis  ?nust  be  zero. 

142.  Def.  Toothed  or  cogged  wheels  are  those  on  the  circum- 
ference of  which  are  projections,  called  teeth  or  cogs.  When 
two  wheels  have  their  cogs  of  such  form  and  distance  that 
those  of  one  will  work  between  those  of  the  other,  the  motion 
of  one  wheel  will  be  communicated  to  the  other  by  the  p?-essure 
of  the  cogs. 


S8 


STATICS. 


When  the  teeth  are  on  the  sides  of  the  wheel  instead  of  the 
circumference,  they  are  called  crown  wheels. 

In  the  preceding  instance,  the  axes  of  the  wheels  and  pin- 
ions are  parallel  or  perpendicular  to  each  other.  When  the 
axes  of  two  wheels  make  an  acute  angle,  the  wheels  take  the 
form  of  frusta  of  cones,  and  are  called  beveled  wheels. 

Axles  on  which  teeth  are  formed  are  called  pinions,  and  the 
teeth  leaves. 

143.  Prop.  Required  the  condition  of  equilibrium  when  the 
action  of  the  power  is  transmitted  to  the  weight  by  means  of 
cogged  wheels. 

Let  S  be  the  mutual 
pressure  of  one  cog  upon 
the  other.  This  pressure 
takes  place  in  the  direction 
of  the  line  S;/z,  m'S  normal 
to  the  cogs  at  the  point  of 
contact, Cm,C'm'  being  per- 
pendiculars from  the  centers 
v->  ^  C  and  C,  of  the  wheels,  on 

that  line. 

Taking  the  moments  about  C,  when  the  power  and  weight 
are  in  equilibrium,  we  have 

P.C'A=S.C'w', 
and  about  C,  W.CB=S.Cm. 

Dividing  the  latter  by  the  former,  we  have 
WC'A  Cm 
P  ~~  CB  'Cm 
Now,  if  the  radii  of  the  axles  are  equal,  or  C'A  =  CB,  we 
shall  have 

WCm 

Y~C'm 

which  gives  the  effect  of  the  action  of  the  cogged  wheels  alone 

or,  since  the  triangles  C'm'o  and  Cmo  are  similar, 

Cm  _Co_W 

TVm'~C'o~  P  * 


WHEEL     AND     AXLE. 


89 


If  the  direction  of  the  line  Sm'  mS  changes  as  the  action 
passes  from  one  cog  to  the  succeeding,  the  point  o  will  also 
change  its  position,  and  the  relation  of  W  to  P  become  va- 
riable. 

But  when  the  cogs  are  of  such  form  that  the  normal  S??i  m'S 
at  their  point  of  contact  shall  always  be  tangent  to  both  circles, 
the  lines  Cm  and  Cm'  will  become  radii,  and  their  ratio  con- 
stant, and  the  point  o  a  fixed  point,  in  which  case 

W_Co_  Cm  _  R 

T~Co~Cm7~R" 

R  and  R'  being  the  radii  of  the  circles  C  and  C  respectively. 

144.  Cor.  When  the  cogs  are  equal  in  breadth,  the  number 
of  cogs  on  C  will  be  to  the  number  of  cogs  on  C  as  the  cir- 
cumference of  C  to  the  circumference  of  "C',  or  as  the  radius 
of  C  to  the  radius  of  C. 

W_number  of  cogs  on  the  wheel  of  W 
P      number  of  cogs  on  the  wheel  of  P  ' 

Scholium.  When  the  working  sides  of  the  cogs  have  the 
form  of  the  involute  of  the  circle  on  which  they  are  raised,  the 
pressure  of  one  cog  on  another  will  always  be  in  the  direction 
of  the  common  tangent  to  the  two  wheels. 

Thus,  let  IHF,  KEb  be  the  two  wheels.  The  acting  face 
GCH  of  the  cog  a  being  formed  by 
the  extremity  H  of  the  flexible  line 
FaH  as  it  unwinds  from  the  circum- 
ference, and  the  acting  face  of  b  by 
the  unwinding  of  the  thread  GE,  the 
line  FCE  will  always  be  normal  to 
the  faces  of  the  cogs  a  and  b  at  their 
point  of  contact.  The  circles  describ- 
ed with  the  radii  AD  and  BD  are  called  the  pitch  lines  of  the 
wheels,  and  will  roll  uniformly  upon  each  other. 

145.  Prop.  Required  the  condition  of  equilibrium  when  the. 
action  of  the  power  is  transmitted  to  the  weight  by  a  system  of 
cogged  wheels  and  pinions. 


90 


STATICS. 


Let  R,  R,,  R2,  &c,  be  the  radii  ol 

the  successive  wheels;  r,rl,r2,&,c.,the 
radii  of  the  corresponding  pinions ;  P, 
Pj,  P2,  &c,  the  powers  applied  to  the. 
circumferences  of  the  successive  wheels. 
Taking  the  moments  about  the  center 
of  each  wheel,  we  have 

P.R=P1r,  P.R^P.r,,  PaR3=Psra,  &c. ; 

since  the  power  applied  to  the  circumference  of  the  second 
wheel  is  equal  to  the  reaction  on  the  first  pinion. 

Multiplying  these  equations  member  by  member,  and  re- 
ducing, 

P„_R.R,.R2!&c. 
P       r.r1.r2,&c. 

or,  the  power  is  to  the  weight  as  the  product  of  the  radii  of  the 

pinions  to  the  product  of  the  radii  of  the  wheels  ; 

or,  as  the  product  of  the  numbers  expressing  the  leaves  of  each 

•pinion  to  the  product  of  the  numbers  expressing  the  cogs  in  each 

wheel. 


EXAMPLES. 

Ex.  1.  If  a  power  of  10  lbs.  balance  a  weight  3f  240  lbs.  on 
a  wheel  whose  diameter  is  4  yards,  required  the  radius  of  the 
axle,  the  thickness  of  the  ropes  being  neglected. 

Ans.  r=3  inches. 

Ex.  2.  The  radius  of  the  wheel  being  2  feet,  and  of  the 
axle  5  inches,  and  the  thickness  of  each  rope  being  £  inch, 
find  what  power  will  balance  a  weight  of  130  lbs. 

Ans.  P=28|  lbs. 

Ex.  3.  The  radius  of  the  wheel  being  3  feet  and  of  the  axle 
3  inches,  find  what  weight  will  be  supported  by  a  power  of 
120  lbs.,  the  thickness  of  the  rope  coiled  around  the  axle  being 
one  inch. 

Ans.  W=1234f. 

Ex.  4.  There  are  two  wheels  on  the  same  axle ;  the  diam- 
eter of  one  is  5  feet,  that  of  the  other  4  feet,  and  the  diameter 


THE     CORD. 


91 


of  the  axle  is  20  inches.  What  weight  on  the  axle  would  be 
supported  by  forces  equal  to  48  lbs.  and  50  lbs.  on  the  larger 
and  smaller  wheels  respectively? 

Ans.  W=264  lbs. 


III. 


THE    CORD. 


146.  The  cord  or  rope  is  employed  as  a  means  of  communi- 
cation of  force.  It  is  regarded  as  perfectly  flexible  and  with- 
out weight,  and  transmits  the  action  of  a  force  applied  at  one 
extremity  to  any  other  point  in  it,  unchanged  in  magnitude,  so 
long  as  it  is  straight,  or  only  passes  over  smooth  obstacles 
without  friction. 

The  force  thus  transmitted  is  called  the  tension. 

Since  the  tension  is  the  same  throughout,  from  one  extremity 
to  the  other,  when  employed  alone,  it  affords  no  mechanical 
advantage  ;  but  when  passed  over  or  attached  to  certain  fixed 
points,  the  resistance  of  these  points  may  be  employed  advan- 
tageously. 

147.  Prop.  Required  the  condition  of  equilibrium  of  a  cord 
acted  upon  by  three  forces. 

Let  the  forces  Pj,  P2, 
P3  be  applied  at  the  ex- 
tremities A  and  B  of  the 
cord  ACB,  and  to  a  knot 
at  C.  Draw  any  line 
CD  in  the  direction  of 
the  force  P2,  and  DN, 
DM  parallel"  to  CA  and 
CB  respectively.  In 
case  ot  equilibrium  {Art. 
28), 

CM  :  CD  :  CN=P1  :  P2  :  P3=  sin.  DCN  :  sin.MCN  :  sin. MCD, 

=  sin.  a'      :  sin.  (a+a')  :  sin.  a  ; 

or,  the  forces  are  each  as  the  sine  of  the  angle  contained  between 
the  directions  of  the  other  two. 


92 


STATICS. 


148.  Cor.  1.  If  the  cord  be  fixed  at  A  and  B,  the  reactions 
of  the  points  A  and  B  take  the  place  of  the  forces  Pj  and  P3, 
and  are  equal  to  the  tensions  of  the  two  parts  of  the  cord  re 
spectively. 

149.  Cor.  2.  If  the  force  P2  be  applied  to  a  running  knot  or 
ring,  the  points  A  and  B  being  fixed,  the  condition  of  equilib- 
rium requires  that  the  direction  of  P„  should  bisect  the  angle 
ACB. 

For  the  point  C  in  its  motion  would  describe  an  ellipse,  A 
and  B  being  the  foci,  and  the  force  P2  could  not  be  in  equilib- 
rium except  when  normal  to  the  curve,  in  which  case  it  bisects 
the  angle  ACB.    Hence,  sin.  a=  sin.  a',  and  Pa=P3 ;  and  since 

Pj  :  P,=  sin.  a' :  sin.  («+a')=  sin.  a  :  sin.  2a=l  :  2  cos.  a. 
.'.  P2=2P,  cos.  a. 

Otherwise  :  since  the  tension  of  the  cord  is  the  same  through- 
out, when  the  cord  passes  over  an  obstacle  without  friction. 
P,=P3 ;  a=a'  and  P2=2P!  cos.  a  {Art.  19). 

150.  Pitor.  Required  the  conditions  of  equilibrium  when  any 
number  of  forces  in  the  same  plane  are  applied  at  different  points 
of  the  cord. 

Let  ABCDE  be  a  cord  to  which  are  applied  the  forces  Pl5 

P2,  P3,  &c,  at  the 
points  A,  B,  C,  &c,  in 
the  directions  AP,, 
BP2,CP3,  &c. 

The  force  P,  at  A 
may  be  considered  as 
acting  at  B  in  its  di- 
rection BA ;  and  since 
B,  when  in  equilibrium, 
must  be  so  from  the 
►action  of  P,,  P2,  ana 
the  tension  of  BC,  the 
resultant  of  P,  and  P2  must  be  in  the  direction  of  CB,  and 
may  be  considered  as  acting  at  C.  Suppose  it  thus  applied, 
and  let  it  be  resolved  into  two,  acting  in  the  directions  Cn  and 
Cm  parallel  to  the  original  components,  and  equal  to  them. 


THE     CORD.  93 

Wo  have  thus  transferred  the  forces  P,  and  iv ,  to  act  at  C 
parallel  to  their  original  directions.  In  the  same  manner,  the 
resultant  of  Pa,  P2,  and  P3  acting  at  C  must  be  in  the  direction 
of  DC,  and  may  be  applied  at  the  point  D  without  disturbing 
the  equilibrium,  and  then  replaced  by  P1?  P2,  and  P3  parallel  to 
their  original  directions,  and  so  on,  for  any  number.  Hence, 
if  all  the  forces  be  supposed  to  act  at  one  point  parallel  to  their 
original  directions,  they  will  be  in  equilibrium.  The  conditions 
of  equilibrium,  therefore,  are  the  same  as  for  any  number  of 
concurring  forces  {Art.  70),  or,  the  sum  of  all  the  forces  re- 
solved in  any  two  rectangular  directions  must  be  zero. 

The  form  which  the  cord  takes  under  the  influence  of  the 
several  forces  is  called  a  funicular  polygon. 

151.  Prop.  Required  the  relations  of  the  forces  which,  acting 
on  a  cord  in  one  plane,  keep  it  in  equilibrium. 

Produce    P2B   to    N, 
P3C  to  M,  &c,  and  let   ; 
the 

Z-ABN=«,  Z.NBC=a', 

z.BCM=i3,  z_MCD=/3', 

&c,  &c. 

Let  tx,  t2,  t3,  &c,  be 
the  tensions  of  the  sev- 
eral successive  portions 
of  the  cord. 

Then  {Art.  147)  t1  :  P2  :  t2=  sin.  a' :  sin.  {a+a')  :  sin.  a;  {a) 
also,  t2  :  P~3  :  t3=  sin.  (3' :  sin.  (j3+j3')  :  sin.  (3,  (a') 

&c,  &c. 

„         ,  .  ,     .  ~        sin.  d  .         _       sin.  a 

From  {a)  we  obtain  <.=P0  - — -. — ; — x,  and  tn=r„- — -. — ; — -^ 
v  '  l        -  sin.  {a+a')  -        -sin.  {a+a') 

sin.  13'  _  sin. /3 

{a')  ^=P3-sin>(i3+Jg,yand^  =  ^3siR<(/3+ig,y 

&c,  &c. 

Equating  the  values  of  t2,  t3,  &c,  we  have 
sin,  a      _  sin.^ 

2Vm.  {a+a')        3"sin.  (P+0')'  W 


94 


STATICS. 


P,.-. 


sin.  /3 


-=P 


sin.  y 


**sin;  (y+y')' 


(b>) 


'sin.  ((3+(3') 
&c, 
for  the  relations  of  the  forces. 

152.  Cor.  1.  If  the  cord  be  fixed  at  the  points  A  and  E,  and 
the  forces  P2,  P3,  P4,  &c,  be  parallel,  we  have 
sin.  a'=  sin.  (3,  sin  ,(3'=  sin.  y,  &c. 
Multiplying  these  equations  by  (b),  (b')  member  by  member 
in  their  order,  we  obtain 


sin.  a  sin.  a'  sin.  j3  sin.  (3' 


?sin.ysm.y' 


But 


'  sin.  (a+a')        3  sin.  ((3+(3')     *  4  sin.  (y+y') 
sin.  a  sin.  a'  sin.  a  sin.  a' 

sin.  (a+a')      sin.  a  cos.  a'+sin.  a'  cos.  a 
1  1 


cos.  a     cos.  a! 
sin.  a     sin.  a' 


cot.  a+cot.  a' 


cot.  a+ cot.  a'     cot./3+cot./3'     cot.  y+cot.  y' 
A      .     N 


:,  &C. 


153.  Cor.  2.  If  the  cord  be 
fixed  at  A  and  E,  and  the  forces 
P2,  P3,  P4,  &c,  be  weights,  the 
horizontal  tension  of  each  por- 
tion of  the  cord  is  the  same. 

For,  by  resolving  the  tension 
of  each  part  horizontally,  wo 
have 

the  horizontal  tension  of  AB=^.  sin.  ABN=£,  sin.  a, 

BC=£2   sin.  BCM=f3  sin.  (3,  &c. 
Substituting  for  tJf  t2,  &c,  their  values  found  above, 

sin.  a  sin.  a!  P„ 


horizontal  tension  of  AB  =  Pn 
BC=P 


3  sin.  (a+a')     cot.  a+  cot.  a1' 
sin.  (3  sin.  j3'_  P3 

sin.  ((3+13')  ~cot.j3+  cot.  (J1 
&c,  &c , 


which,  by  Co?:  1,  are  all  equal. 


THE     CORD. 


95 


154.  Cor.  3.  Since  the  reactions  of  the  points  A  and  E  equi- 
librate the  resultant  of  all  the  weights,  the  lines  AB,  ED  pro- 
duced, must  meet  in  some  point  of  the  vertical  through  the 
center  of  gravity  of  the  system.  For  three  forces,  if  in  equi- 
librium, must  meet  in  a  point. 

155.  A  heavy  cord  may  be  considered  a  funicular  polygon 
loaded  with  an  infinite  number  of  small  weights,  and  since  the 
number  of  weights  is  infinite,  the  polygon  will  also  have  an 
infinite  number  of  sides,  or  will  become  a  curve. 

The  curve  which  a  heavy  cord  or  chain  of  indefinitely  small 
links  will  assume,  when  suspended  from  two  fixed  points  not  in 
the  same  vertical  line,  is  called  the  catenary. 


EXAMPLES. 

Ex.  1.  Two  equal  weights  balance,  by  a  cord,  over  any 
number  of  fixed  points  without  friction.  Required  the  press- 
ure on  each. 

Since  the  tension  of  the  cord  PjABCDEP£ 
throughout,  each  point  is  acted  upon 
by  three  forces,  viz.,  two  equal  ten- 
sions on  each  side  of  it  and  the  reac- 
tion of  the  point,  which  last  must  be 
equal  to  the  resultant  of  the  other  two. 
Hence,  calling  the  angles  at  A,  B,  C, 
&c,  a,  b,  c,  &c,  by  Art.  15,  and  Cor., 
Art.  19, 

the  pressure  on  A=2PT  cos.  ^a, 
B=2P1  cos.l&, 
&c,  &c. 

Ex.  2.  A  cord  of  given  length  passes  over  two  fixed  points 
A  and  B  without  friction,  and  one  extremity,  to  which  a  given 
weight  P  is  attached,  passes  through  a  small  ring  at  the  other 
extremity  C.  It  is  required  to  find  the  tension  of  the  cord  when 
in  equilibrium,  and  the  length  of  the  part  CP  below  the  ring. 

Since  the  cord  passes  freely  over  the  points  A  and  B,  and 
through  the  ring  C,  it  is  of  the  same  tension  throughout,  and 
equal  to  the  weight  P.     Hence  the  point  C  is  kept  at  rest  bv 


96 


STATICS. 


,B  three  equal  forces,  and,  by  Art.  18,  must  make 
angles  of  120°  with  each  other.  Draw  the 
horizontal  line  AD;  and,  since  ACD— 120°, 
ADC  and  CAD  are  each  equal  to  30°.  The 
position  of  A  and  B  being  given,  the  angle 
BAD  must  be  known.  Hence,  in  the  triangle  ACB  we  have 
the  side  AB  and  all  the  angles  from  which  AC  and  BC  may 
be  determined.  Then  the  whole  length  of  the  cord,  diminished 
by  the  perimeter  of  the  triangle,  will  be  the  distance  of  the 
weight  from  the  ring. 


§   IV.    THE    PULLEY. 

156.  The  pulley  is  a  small  grooved  wheel  movable  about 
an  axis,  and  fixed  in  a  block.  The  cord  passes  over  the  cir- 
cumference of  the  wheel  in  the  groove. 

The  use  of  the  pulley  is  to  prevent  the  effects  of  friction  and 
rigidity  of  the  cord.  The  first  of  these  it  diminishes  by  trans- 
ferrins; the  friction  from  the  cord  and  circumference  of  the 
wheel  to  the  axle  and  its  supports,  which  may  be  highly  pol- 
ished or  lubricated.  The  effects  of  rigidity  are  diminished  by 
turning  the  cord  in  a  circular  arc  instead  of  a  sharp  angle. 

The  pulley  is  called  fixed  or  movable,  according  as  the  block 
is  fixed  or  movable. 


,  "         IT7"    '    -• 


B| 


157.  The  fixed  pulley  serves  merely  to 
change  the  direction  of  the  forces  trans- 
mitted by  the  cord,  since,  neglecting  the 
friction  of  the  pulley,  the  tension  of  the 
cord  is  the  same  in  every  part  of  it.  Hence 
the  power  equals  the  weight,  and  the  press- 
ure on  the  axis  of  the  pulley  equals  their 
sum. 


THE     PULLEY. 


9-? 


158.  Prop.  Required  the  relation  of  the  power  to  the  weight  m 
the  single  movable  pulley. 

The  tension  t  of  the  cord,  being  the  same  p 
throughout,  is  equal  to  the  power  P ;  also 
to  the  pressure  on  the  hook  Q. 

The  resultant  of  the  two  tensions  in  the 
directions  CP  and  DQ,  being  equal  and 
opposite  to  the  weight  W,  must  be  vert- 
ical. Let  a  be  the  angle  made  by  the 
cords  with  this  vertical,  and,  resolving  the 
tensions  vertically,  we  have 

2t  cos.  a=2F  cos.  a—W. 

W 
.'.   P=: 


2  cos.  a' 
the  same  as  obtained  in  Art.  149. 

Cor.  1.  If  the  weight  w  of  the  pulley  be  taken  into  account, 

P= . 

2  cos.  a 

Cor.  2.  If  the  cords  are  parallel,  a=0  and 

W+m; 
•    Jl —       z — . 


Cor.  3.  If  a=90°,  2a=180°,  or  the  cord  becomes  straight 
and  horizontal.     In  this  case 

n     W+w 

or  the  power  must  be  infinite.  In  other  words,  no  power  can 
reduce  the  cord  to  a  horizontal  straight  line  while  the  weight 
is  finite. 

159.  Of  the  various  combinations  of  pulleys  there  are  three, 
which  we  shall  distinguish  by  the  first,  second,  and  third  sys- 
tems of  pulleys. 

160.  Prop.  Required  the  relation  of  the  power  to  the  weight  in 
the  first  system  of  pulleys. 

G 


98 


STATICS. 


The  annexed  figure  represents  this  system. 
Neglecting  friction,  the  tension  t  of  the  cord 
is  the  same  throughout,  and  equal  to  P.  The 
weight  W  and  the  weight  of  the  lower  block 
w  are  sustained  by  the  tensions  of  the  several 
cords  at  the  lower  block.  Hence,  if  n  be  the 
number  of  cords  at  this  block, 
nt=nP=W+w 

p=W+to 
n 

If  the  weight  of  the  lower  block  be  neg- 
lected, 

n 
Of  this  system  of  pulleys  there  are  various 
modifications.     The  annexed  form  is  the  one 
in  most  common  use. 


161.  Prop.  Required  the  relation  of  the  power  to  the  weight  in 
the  second  system  of  pulleys. 

The  annexed  figure  represents  this  system  with  three  mov- 
able pulleys,  each  pulley  having  its  own  rope. 
Designate  by  a,,  a2,  a3,  &c,  the  pulleys  re- 
spectively in  their  order  from  the  weight  W, 
by  wx,  w2,  iv 3,  &c,  their  weights,  and  by  tlt 
t2,  t2,  &c,  the  tensions  of  their  respective 
cords. 

Then,  for  the  equilibrium  of  a,,  we  have 

W+10. 
2t1=W+w1,  or  *,= -— K 


For  the  equilibrium  of  a2,  2t2=t1  + 


v: 


_W+wl+2w, 


or 


to  = 


W+u>,+2ios 


For  the  equilibrium  of  c  3,2t3=t2  + 


to. 


W+w^+2w2+2'u>i 


or 


THE     PULLEY. 

W+w1+2w2+2*w, 


99 


For  the  equilibrium  of  an. 

W+io1+2M>3+29to3+2'M>4+  .  .  .  2"-'mv. 


P=tn 


2" 


Cor.  1.  If  the  weight  of  each  pulley  is  the  same,  and  equal  w 
^     W+w1(l+2+22+23+....2-1) 


_W     10,(2"-!) 
~2n+        2" 


W 


1 


Cor.  2.  If  the  weights  of  the  pulleys  be  neglected, 

W 
P=_,  or  W=2".P. 

162.  Prop.  Required  the  relation  of  the  power  to  the  weig.u 
in  the  third  system  of  pulleys. 

In  this  system  each  cord  is  attached  to  the 
weight,  and  the  number  of  movable  pulleys  is  one 
less  than  the  number  of  cords.  Designating  the 
pulleys  in  their  order  from  the  weight  by  ax,  a2, 
a3,  &,c,  their  weights  respectively  by  iol,  w2, 
u)3,&c,  and  the  tensions  of  the  successive  cords 
by  £j,  t2,  t3,  &c,  we  have 

t2=2t1+w1=2V+w1, 
t3=2ta+wa=2,l?+2w1+u^, 

ti=2t3+w3=2'V+2''w1+2w2  +w3. 


And  if  there  be  n  cords, 

tn=2"-1V+2n-hol+2"-'w2+  &c 2wn^+u\^. 

But  W=t,+t2+t3+,&tc U 

=P(1  +2+22+23+,  &c. . . .  2"-1)  +iv ,  (1 +2+22+,  &c 
.  .  .  2'-2)+u;2(l+2+22+.&c 2"-3)-h&c. 


100 


STATICS. 


=  P(2"-l)  +  u>1(2n-1-l)+w2(2"-:-l)+,&c 

w„_1(2-l). 
If  the  (ft— 1)  pulleys  are  of  the  same  weight  u>,, 

W=P(2n-l)+w1(2n-1+2-2+2'-3+,&c 2"+2  +  l-n) 

='P(2n-l)+w1(2n-l)-iu1n 
=  (Y+wl)(2n-l)-iv1n. 
If  the  weights  of  the  pulleys  be  neglected, 
W=P(2"-1). 


EXAMPLES. 


Ex.  1.  At  what  angle  must  the  cords  of  a  single  movable 
pulley  be  inclined  in  order  that  P  may  equal  W  ? 

Ex.  2.  In  the  first  system  of  pulleys,  if  there  be  10  cords  at 
the  lower  block,  what  power  will  support  a  weight  of  1000  lbs.  1 

Ex.  3.  In  the  second  system  of  pulleys,  if  1  lb.  support  a 
weight  of  128  lbs.,  required  the  number  of  pulleys  supposed 
without  weight. 

Ex.  4.  In  the  third  system  of  6  pulleys,  each  weighing  1  lb., 
find  what  weight  will  be  supported  by  a  power  of  12  lbs 


Ex.  5.  Find  the  ratio  of  the  power  to  the 
weight  in  the  annexed  modification  of  the  second 
system  of  pulleys. 


§  V.    THE    INCLINED    PLANE. 

163.  The  Inclined  plane,  as  a  mechanical  power,  is  supposed 
perfectly  hard  and  smooth,  unless  friction  be  considered.  It 
assists  in  sustaining  a  heavy  body  by  its  reaction.  This  reac- 
tion, however,  being  normal  to  the  plane,  can  not  entirely 
counteract  the  weight  of  the  body,  which  acts  vertically  down- 


THE     INCLINED     PLANE. 


101 


ward.  Some  other  force  must  therefore  be  made  to  act  upon 
the  body,  in  order  that  it  may  be  sustained. 

164.  Pkop.  Required  the  conditions  of  equilibrium  of  a  body 
sustained  by  any  force  on  an  inclined  plane. 

Let  AB  be  a  section  of  an  inclined  plane,  of  which  AB  is 
the     length,     BC     the  P 

height,  and  AC  the 
base.  Let  i  be  the  in- 
clination of  the  plane  to 
the  horizon,  e  the  angle 
made  by  the  direction 
of  the  power  P  with  the 
plane  AB,  W  =  the 
weight  of  the  body  ay 
and  R=  the  reaction  of  "W~V 

the  plane.  The  body  is  kept  at  rest  by  the  action  of  P,  W, 
and  R.  Resolving  the  forces  parallel  and  perpendicular  to 
the  plane,  we  have 


Pcos.  e— W  sin.  ?'=0, 
R+P  sin.  e— W  cos.  i=0. 
From  (a)  we  obtain 

W       COS.  £ 


P 


sin.  7 


From  (b),  R=W  cos.  i— P  sin.  £=W  cos.  z- 


(a) 


(P) 

W  sin.  i  sin.  e 


COS.  £ 


W. 


cos.  (i+£) 


W 


COS.  £ 
COS.  £ 


or 


(d) 


R      cos.  (i+e)' 
the  same  relations  as  obtained  in  Ex.  24,  Art.  129. 

165.  Cor.  1.  If  the  force  P  act  parallel  to  the  plane,  £-0 
and  (c)  becomes 

W_    1       AB 
P_sin.2~BC' 
or,  the  power  is  to  the  iveight  as  the  height  of  the  plane  to  iti 
length. 


102 


STATICS. 


o  W       1        AB 

From(^veget-r--^, 

or,  the  reaction  of  the  plane  is  to  the  weight  as  the  base  to  the 
length. 

166.  Cor.  2.  If  the  power  act  parallel  to  the  base  of  the  plane, 
f— — if  and  (c)  becomes 

W    cos.  i    AC 

T==sin~£==BC' 

or,  the  power  is  to  the  weight  as  the  height  to  the  base. 

W    cos./    AC 
From  (d),  -^ =— =AB' 

or,  the  reaction  of  the  plane  is  to  the  weight  as  the  length  to  the 
base. 

167.  Prop.  Required  the  conditions  of  equilibrium  of  two 
bodies  resting  on  two  inclined  planes  having  a  common  summit, 
the  bodies  being  connected  by  a  cord  passing  over  a  pulley  at 
the  summit. 


Let  W  and  W,  be  the 

weights  of  the  bodies,  and 
i,i,  the  inclinations  of  the 
planes. 

If  t  be  the  tension  of  the 
cord,  we  have  for  equilib- 
rium   on   the    plane    AB 
c    {Art.  165), 


t=W.  sin.  i, 
on  the  plane  BC  <=W,  sin.  £I. 

.*.  Wsin.  i=Wj  sin.  ff, 

AB      VlBC 

WAB 

W^BC' 

or,  the  weights  are  proportional  to  the  lengths  of  the  plana  on 
which  they  rest  respectively. 


or 


or 


THE     WEDGE. 


103 


EXAMPLES. 


Ex.  1.  What  force  acting  parallel  to  the  base  of  the  plane 
is  necessary  to  support  a  weight  of  50  lbs.  on  a  plane  inclined 
at  an  angle  of  15°  to  the  horizon  ? 

Ex.  2.  If  the  weight,  power,  and  reaction  of  the  plane  are 
respectively  as  the  numbers  25,  16,  and  .10,  find  the  inclination 
of  the  plane,  and  the  inclination  of  the  power's  direction  to 
the  plane. 


§  VI. 


TEIE    WEDGE. 


168.  The  wedge  is  a  triangular  prism  whose  perpendicular 
section  is  an  isosceles  triangle.  The  dihedral  angle  formed 
by  the  two  equal  rectangular  faces,  is  called  the  angle  of  the 
wedge.  The  other  rectangular  face  is  called  the  back.  It  is 
used  to  separate  the  parts  of  bodies,  by  introducing  the  angle 
of  the  wedge  between  them  by  a  power  applied  perpendicu- 
larly to  the  back.  The  equal  rectangular  faces  are  regarded 
as  perfectly  smooth,  in  which  case  the  only  effective  part  of 
the  resistance  must  be  perpendicular  to  these  faces. 

169.  Prop.  To  determine  the  conditions  of  equilibrium  in  the 
wedge. 

Let  ABC  be  a  section  of  the 
wedge  perpendicular  to  the  angle 
or  edge  A.  Draw  AD  bisecting 
the  angle,  and  let  BAD=CAD=a. 
Let  2P  be  the  power  applied  to  the 
back  BC  of  the  wedge,  which  must 
be  in  equilibrium  with  the  pressures 
R  on  the  two  faces  AB  and  AC. 
If  an  equilibrium  exist,  the  forces 
2P,  R,  and  R  must  meet  at  some 
point  in  AD  {Art.  74). 

Resolving  the  forces  vertically, 
or  in  the  direction  AD,  we  have 


104 


STATICS. 


or 


or 


P       . 

Y7=  sin.  a 
K 


2R.sin.BAD-2P=0, 
P=R.  sin.  a, 
BD    BD.Z    \  the  back  of  the  wedge 


B A    BA.Z  face  of  the  wedge 

where  I—  the  length  of  the  edge  or  breadth  of  the  face. 

170.  In  the  foregoing  investigation  of  the  theory  of  equilib- 
rium in  the  wedge,  we  have  omitted  the  consideration  of  the 
friction,  and  have  supposed  the  power  to  be  a  pressure ; 
whereas,  in  practice,  the  wedge  is  kept  at  rest  by  friction  alone, 
and  the  power  arises  from  percussion.  The  following  prob- 
lem will  serve  to  elucidate  the  theoretical  view  here  taken  of 
the  wedge. 

171.  Prob.  A  heavy  beam  is  attached,  by  a  hinge  at  one 
end,  to  a  smooth,  horizontal  plane,  while  the  other  rests  on  the 
smooth  face  of  a  semi-wedge.  Required  the  horizontal  force 
necessary  to  keep  the  wedge  from  moving. 

Let  DE  be  the  beam 
and  BAC  the  wedge, 
Z.BAC  =  a,  z.ADE  =  /3, 
/=  the  length  of  the 
beam,  g  the  center  of 
D  |1V  A  33  gravity,  and  Dg=a. 

The  wedge  is  kept  in  equilibrium  by  the  pressure  of  the 
beam  upon  it  at  E,  and  the  horizontal  force  P  acting  upon  it 
at  some  point  H.  The  beam  is  kept  in  equilibrium  by  its 
weight  w  acting  at  g  and  the  reaction  R  of  the  face  of  the 
wedge  at  E. 

Taking  the  moments  about  D  for  the  equilibrium  of  the  beam, 
we  avoid  expressions  involving  the  unknown  thrust  R',  and 
have 

w.Dg.  cos.  ADE-R.DE.  sin.  DER=0, 
or  w.a.  cos.  j3— R.Z.  cos.  (a— 13)=0. 

a      cos.  (3 
/  cos.  {a—li) 

But  the  principles  of  the  wedge  give 
P=R.  sin.  a. 


THE     SCREW. 


105 


_         a  sin.  a  cos.  /3 

..    P  =  tt?.y. — . 

/  cos.  (a—fj) 
By  an  examination  of  this  value  of  P,  it  will  be  seen  that  the 
power  necessary  to  keep  the  wedge  from  moving  will  diminish 
as  the  wedge  advances  beneath  the  beam. 


//£."' 


§  VII.    THE    SCREW. 

172.  If  we  divide  the  rectangle  ABCD  into  equal  parts  by 
the  lines  mn,  m'n',  &c,  parallel 
to  AB,  and  draw  the  parallel  di- 
agonals of  the  rectangles  thus  m 
formed,  and  if  we  suppose  the 
whole  rectangle  to  be  wrapped 
round  the  surface  of  a  cylinder, 
the  perimeter  of  whose  base  is 
equal  to  AB  and  altitude  to  BC, 
the  diagonals  of  the  rectangles 
will  trace  on  the  surface  of  the  cylinder  a  continuous  curve, 
which  is  called  the  helix. 

If  a  projecting  thread  or  rib  be  attached  to  the  cylinder  upon 
this  curve,  we  have  the  screw,  sometimes  called  the  external 
screw.  Similarly,  if  we  take  a  hollow  cylinder  of  exactly  the 
same  radius  as  the  solid  one,  and  generate  a  groove  in  the 
same  curve,  we  have  the  internal  screw  or  nut. 

The  screw  works  in  the  nut,  either  of  which  may  be  fixed 
and  the  other  movable. 

173.  Prop.  To  determine  the  conditions  of  equilibrium  in  the 
screw. 

From  the  construction  of  the 
screw,  it  appears  that  the  thread 
of  it  is  an  inclined  plane,  of 
which  the  base  is  the  circum- 
ference of  the  cylinder  and  the' 
height  the  distance  between  the 
threads.  The  force  is  generally 
applied  perpendicularly  to  the 
end  of  a  lever  inserted  into  the 


1 06  STATICS. 

cylinder,  and  in  the  plane  perpendicular  to  ti\e  axis  of  the 
cylinder.  The  power  P  thus  applied,  in  turning  the  screw 
round,  produces  a  pressure  on  the  threads  of  the  screw  in  the 
direction  of  the  axis  of  the  cylinder.  In  case  of  equilibrium, 
let  the  counterpoise  of  this  pressure  be  W.  Let  ABC  be  the 
inclined  plane  formed  by  unwrapping  one  revolution  of  a 

thread,  and  let  w— -th  part  of  W,  be  supported  at  a,  q  being 

the  same  part  of  Q,  the  force  applied  at  the  circumference  of 
the  cylinder.  Put  FD  =  a,  ED  the  radius  of  the  cylinder  =r, 
and  angle  BAC=a.     Then,  Art.  133, 

«-*^"-*£-  » 

By  Art.  166,  V=W^X£> 

Q,     W        distance  of  two  threads 


or 


n      n  '  circumference  of  the  cylinder' 

and  the  same  holds  for  each  of  the  other  portions  of  W  at  the 

other  points  of  the  plane.     Therefore,  we  have 

_w        distance  of  two  threads 

'  circumference  of  the  cylinder' 

t-.     ,  .    ^     -^  27ra     _  circumference  described  by  P 

By  (a),  Q=P.-— =P. : ? e — Fj    — • 

J   v  2nr  circumierence  ot  cylinder 

W      circumference  described  by  P 
P  —        distance  of  two  threads      ' 
or,  the  power  is  to  the  weight  as  the  distance  of  two  threads 
is  to  the  circumference  described  by  the  power  in  one  revolu- 
tion of  the  screw. 

It  will  be  seen  that  the  ratio  of  P  to  W  is  independent  of 
the  radius  of  the  cylinder. 

EXAMPLES. 

Ex.  1.  What  force  must  be  exerted  to  sustain  a  ton  weight 
on  a  screw,  the  thread  of  which  makes  150  turns  in  the  height 
of  1  foot,  the  length  of  the  arm  being  6  feet  ? 

Ex.  2.  Find  the  weight  that  can  be  sustained  by  a  powei 


BALANCES,    ETC. 


10*3 


of  1  lb.  acting  at  the  distance  of  3  yards  from  the  axis  of  the 
screw,  the  distance  between  the  threads  being  1  inch. 

Ex.  3.  What  must  be  the  length  of  a  lever  at  whose  ex- 
tremity a  force  of  1  lb.  will  support  a  weight  of  1000  lbs.  on  a 
screw,  whose  threads  are  f  inch  apart? 


§   VIII.    BALANCES    AND    COMBINATIONS    OF    THE    MECHANICAL 
POWERS. 

174.  The  common  balance,  in  its  best  form,  is  a  bent  lever, 
in  which  the  weight  of  the  lever  must  be  taken  into  consider- 
ation. 

In  the  annexed  figure  the  points  A  and  B,  from  which  the 
scale-pans  and  weights 
are  suspended, are  call- 
ed the  points  of  suspen- 
sion ;  C  is  the  fulcrum, 
being  the  lower  edge 
of  a  prismatic  rod  of 
steel  projecting  on  each 
side  of  the  beam;  when 
the  balance  is  in  use, 
these  edges  on  each 
side  of  the  beam,  as  at 
C,  rest  on  hard  surfa- 
ces, so  that  the  beam  turns  freely  about  C  as  a  fulcrum. 

175.  The  requisites  of  a  good  balance  are, 

1°.  That  the  beam  rest  in  a  horizontal  position  when  loaded 
with  equal  weights.  , 

2°.  That  the  balance  possess  great  sensibility. 
3°.  That  it  possess  great  stability. 

176.  Prop.  To  determine  the  conditions  that  the  beam  rest  in 
a  horizontal  position  when  loaded  with  equal  weights. 

Supposing  the  beam  horizontal,  in  case  of  equilibrium,  if  we 
neglect  the  weight  of  the  beam,  the  moments  of  the  weights 
must  be  equal,  and,  therefore,  the  arms  must  be  equal  {Art.  133). 


108 


STATICS. 


But  taking  into  consideration  the  weight  of  the  beam,  its  cen 
ter  of  gravity  must  be  in  the  vertical  through  C,  the  center  of 
motion  {Art.  110),  and,  in  order  to  this,  the  beam  must  be  sym- 
metrical on  opposite  sides  of  the  fulcrum  {Art.  97).  The  line 
AB,  joining  the  points  of  suspension,  is  obviously  bisected  by 
the  vertical  through  C  and  the  center  of  gravity  of  the  beam, 
and  the  point  of  intersection,  for  reasons  which  will  appear, 
should  be  below  C. 

177.  Prop.  To  determine  the  conditions  that  the  balance  may 
possess  great  sensibility. 

Let  C,  A,  and  B  be  the  fulcrum  and  points  of  suspension,  as 

-y  in  the  preceding  figure,  and 
join  Cg,  the  centers  of  mo- 
■N  tion  and  of  gravity.  Cg  is 
perpendicular  to  AB  and  bi- 
sects it,  if  the  beam  is  con- 
structed in  accordance  with 
the  first  requisite.  Let  M 
C  E  N  be  a  horizontal  line 
through  C,  meeting  AB  in  E 
making  with  it  an  angle  equal  to  6. 

The  sensibility  is  measured  by  the  amount  of  deflection  8 
of  the  line  AB  from  a  horizontal  position  by  a  given  small  dif- 
ference P— Q,  of  the  weights. 

Draw  the  vertical  lines  Dd  and  ga,  and  put  AD=BD~a, 
CD=d,  Cg~h,  and  weight  of  the  beam  =iv.  Now  MN  is  bi- 
sected in  d,  and  Md=a  cos.  0,  Cd=d.  sin.  6,  and  Ca=h.  sin.  6. 

If  the  system  is  in  equilibrium,  the  moments  about  C  give 

P.CM-Q.CN-u>.Ca=0, 
or  P.{M.d-Cd)-Q,(Nd+Cd)-w.Ca=0, 

or  (P-Q)a.  cos.  0.-(P+Q)tfsin.  6-w.h.  sin.  0=0, 

or  (P-Q)a-{(P+Q).rf+u>./i}tan.0=O. 

.    tan  P-     (P~Q)-g 

Hence  the  angle  6,  and,  therefore,  the  sensibility,  is  increased 
for  given  values  of  P  and  Q,  by  mc?*easing  the  lengths  of  the 


BALANCES,    ETC.  109 

arms  (a),  by  diminishing  the  weight  of  the  beam  (w),  or  by  di- 
minishing the  distances  of  the  fulcrum  from  the  center  of  grav- 
ity of  the  beam  (Ji)  and  from  the  line  joining  the  points  of  sus- 
pension (d). 

178.  Prop.  To  determine  the  conditions  that  the  balance  may 
possess  great  stability. 

If  the  balance  be  loaded  with  equal  weights  and  disturbed 
from  its  position  of  equilibrium,  the  rapidity  with  which  it  re- 
turns to  that  position  is  a  measure  of  its  stability.  But  this 
rapidity  of  return  to  a  horizontal  position  will  depend  upon  the 
moment  which  urges  it  back. 

But  this  moment  is,  since  P=Q, 

P.CN-P.CM+w.Ca, 

or  ¥.2.Cd+w.Ca, 

or  (2P.d+w.h)  sin.  0. 

Hence,  for  given  values  of  P  and  d  the  stability  is  greater, 
as  d,  h,  and  w  are  increased. 

179.  Cor.  Hence,  by  increasing  the  stability,  we  diminish 
the  sensibility,  but  the  sensibility  may  be  increased  by  increas- 
ing the  length  of  the  arms,  without  affecting  the  stability. 

For  commercial  purposes,  when  expedition  is  required,  ana 
the  material  weighed  is  not  of  great  value,  sensibility  is  sacri- 
ficed to  stability ;  but  for  philosophical  purposes  great  sensi- 
bility is  required,  and  stability  is  of  little  comparative  import- 
ance. 

THE    STEELYARD    BALANCE. 

180.  The  steelyard  balance,  or  Roman  steelyard,  is  a  level 
of  the  first  kind,  with  un- 
equal arms.     The  body 

W  to  be  weighed  is  hung         \^j|Y  \    f=§    §  j 
at  the  shorter  arm  A,  and        j*?  p 

a  given  constant  weight 
P   is   moved    along   the 
other  arm  till  it  balances  W;  then  the  weight  of  W  is  known 
from  the  place  of  the  counterpoise  P. 


W 


110 


STATICS. 


181.  Prop.    To  determine  the  law  according  to  which   th 
longer  arm  of  the  steelyard  must  be  graduated. 

Let  G  be  the  center  of  gravity  of  the  beam  AB  and  w  its 
weight.  Put  OA=jo,  OC=pl,  OG=g,  and  taking  moments 
about  the  fulcrum  O,  in  case  of  equilibrium  we  .have 

Wp  —  Wg—  Pj9  j  =  0. 

wg    Wp 
Taking  Ox,  a  fourth  proportional  to  P,  w,  and  g,  we  have 

Hence  the  distance  from  x  to  the  counterpoise  P  varies  as 
the  weight ;  and  if  the  weights  be  in  arithmetical  progression, 
the  distances  xl,  x2,  x3,  &c,  will  also  be  in  arithmetical  pro- 
gression. 

THE    BENT    LEVER    BALANCE. 


182.  This  balance  is  represented  b\ 
the  annexed  figure,  where  ABC  is  the 
bent  lever  turning  about  a  pivot  at  B. 
A  scale  (E)  hangs  from  A,  and  at  C  an 
index  points  to  some  division  on  the 
graduated  arc  GCF. 


183.  Prop.  To  determine  the  principle  of  graduation  in  the 
bent  lever  balance. 

Let  g  be  the  center  of  gravity  of  the  beam  at  which  its 
weight  w  acts.  The  weight  of  the  scale  (E)  and  the  weight 
(W)  of  a  body  placed  in  it  will  act  vertically  through  A.  Let 
the  horizontal  line  through  B  meet  the  vertical  lines  through 
g  and  A  respectively  in  D  and  K.  Then  the  moments  about 
B,  in  equilibrium,  give 

w.DB-(E+W)BK=0.  (a) 


BALANCES,     ETC. 


Ill 


As  greater  weights  are  put  into  the  scale  E,  the  point  A  ap- 
proaches more  nearly  the  vertical  through  B  from  the  bent 
form  of  the  beam,  or  BK  diminishes,  while  BD  increases. 

Suppose  the  point  A  to  be  at  K  when  the  scale  is  unloaded, 
and  let,  in  this  position  of  the  beam,  the  angle  DBg=6.  When 
a  weight  is  put  into  the  scale  the  point  A  will  descend  through 
some  angle  0,  and  the  arm  Bg  will  rise  through  the  same  angle. 
In  this  new  position  the  angle  DBg-  will  become  0—0.  Let 
Bg=a  and  BA=b ;  then 

DB  —  a  cos.  (0—0),  and  BK=&  cos.  0. 
These  substituted  in  (a),  give 

w.a.  cos.  (6— 0)  —  E.&.  cos.  0— W.b.  cos.  0=0, 
or  w.a.  cos.  0  cos.  (p-\-w.a  sin.  6  sin.  0— E.5  cos.  0— W.b.  cos.  0=0, 
or  w.a.  cos.  6+w.a  sin.  6  tan.  0  —  E.&— W.b—0. 


tan.  (p- 


W.b 


;+■ 


E.6  —  w.acos.d 


w.a.  sin.  u         w.a.  sin.  0 

Hence,  tan.  0  varies  as  W,  and  the  limb  GF  must  be  di- 
vided into  arcs  whose  tangents  are  in  arithmetical  progression. 

Practically,  the  limb  may  be  graduated  from  the  positions 
of  the  index  at  C  for  a  succession  of  weights  put  into  E.  This 
instrument  possesses  great  stability. 


ROBERVAL  S    BALANCE. 


184.  This  instrument  is  of  greater  interest  from  its  paradox- 
ical appearance  than  from  its  use  as  a  machine  for  weighing 
bodies.  Its  discussion  affords  an  interesting  application  of  the 
doctrine  of  couples. 

It  consists  of  an  upright 
stem  upon  a  heavy  base  A, 
with  equal  cross-beams 
turning  about  pivots  at  a 
and  b.  These  cross-beams 
are  connected  by  pivots  at 
c,  d,  e,  and/,  with  two  oth- 
er equal  pieces  in  the  form 


Li 


M 


A 


112 


STAT'  C  S. 


weights  are  suspended  from  the  horizontal  arms 


ofaT.     The 

of  the  latter  pieces 

185.  Prop.  In  RobervaVs  balance,  equal  weights  balance  at  all  i 
distances  from  the  upright  stem. 

Let  the  letters  in  the 
annexed  figure"  indi- 
cate the  same  parts  as 
in  the  former. 

Let  equal  and  oppo- 
site forces  P,  and  P2 
each  equal  to  P  act  in 
ec,  and,  similarly,  let 
P3  and  P4   act  in  df. 


Q2<^ 


These  forces  Pa,  P. 
P„  and  P4 


do  not  dis- 

Lurb  the  equilibrium.  Now  P  at  B,  as  in  Art.  71,  is  equivalent 
to  P,  at  e  and  the  couple  P,  B/j,  P2,  and,  similarly,  P  at  C  is 
equivalent  to  P4  at  /  and  the  couple  P,  Ck,  P3 ;  and  since  P, 
at  e  balances  P4  at  f,  we  have  only  the  two  couples  to  dis- 
pose of. 

Now  for  the  couple  P,  Bh,  P2,  Arts.  58  and  59,  we  may  sub- 
stitute a  couple  Q,,,ec,  Q2  in  its  own  plane  and  of  equal  mo- 
ment, in  which  the  forces  Qx  and  Q2,  acting  in  the  directions 
of  the  cross-beams  cd  and  ef  (which  always  remain  parallel  to 
each  other  as  they  turn  on  the  pivots  a  and  b),  are  destroyed  by 
the  resistance  of  the  pivots  a  and  b.  Similarly,  the  couple  P, 
Ck,  P3  may  be  replaced  by  the  equivalent  couple  K1,fd,  R2, 
in  which  Rj  and  R2  are  destroyed  by  the  resistance  of  &  and 
a.  These  new  couples,  therefore,  do  not  disturb  the  equilibri- 
um, and  the  original  forces  P  at  B  and  P  at  C  must  be  in  equi- 
librium. 

If  the  beams  cad  and  ebf  be  moved  round  the  pivots  into 
any  oblique  position,  the  same  reasoning  would  apply,  and  the 
equilibrium  still  subsist. 

Cor.  Unequal  weights  can  not  balance  from  whatever  points 
suspended 


BALANCES,     ETC. 


n& 


B 


C" 


B" 


W 


B' 


"W" 


A"' 


c 

_S7 


>T 


A" 


W" 


186.  Prop.   To  determine  the  ratio  of  the  power  to  the  weight 
in  a  combination  of  levers. 

Let  the  power  P  act  at  A, 
and  the  weight  W  at  B'". 
The  first  three  levers  are  of 
the  second  kind,  and  the  last 
one  of  the  first  kind,  the  ful- 
crums  being  at  C,  C',  C",  and 
C".  Let  BA',  B'A",  B"A'" 
be  rigid  rods  connecting  the 
levers,  and  let  the  action  of 
the  first  lever  on  the  rod  BA' 
be  W,  which  becomes  the 
power  acting-  on  the  second, 
W"  and  W"  the  weights  to  the  second  and  third  respectively, 
and  powers  to  the  third  and  fourth. 

P     CB    W     C'B'    W"     C"B"        ,  W"    C"B' 


a" 


B' 


ThenTT77=; 


C"B"        ,  W 
— ,  and 


W    CA'  W"    C'A"  W"    C"A"'  "~"    W      C'"A' 
Taking  the  continued  product  of  these  equations  member 
Dy  member,  we  have 

P  _  CBxC'B'xC"B"xC'"B'" 

W-CAxC'A'xC"A"xC"'A"/ ' 

or,  the  ratio  of  the  power  to  the  weight  in  the  combination  is 
equal  to  the  product  of  the  ratios  of  the  power  to  the  weight  in 
each  lever. 

187.  Prop.   To  determine  the  ratio  of  the  power  to  the  weight 
in  the  endless  screw. 


This  machine  is  a  combination  of  the 
screw  and  wheel  and  axle. 

Let  P  be  the  power  applied  to  the  han- 
dle of  the  winch,  W  the  pressure  of  the 
screw  on  the  teeth  of  the  wheel,  and  W 
the  weight  suspended  from  the  axle  of 
the  wheel.     Then 


H 


and 


114  STATICS. 

P      distance  between  the  threads  of  the  screw 
W        circumference  described  by  the  power    ' 
W'_  radius  of  axle 
VV     radius  of  wheel' 
P     distance  between  two  threads  of  screw 
W     circumference  described  by  the  power 
radius  of  axle 

V  ; • 

radius  of  wheel ' 
or,  the  ratio  of  the  power  to  the  weight  in  the  endless  screw  is 
equal  to  the  product  of  the  ratios  of  the  power  to  the  weight  in 
the  screw  and  in  the  wheel  and  axle. 

188.  Prop.  To  determine  the  ratio  of  the  power  to  the  weight 
in  any  combination  of  the  mechanical  powers. 

Let  P=  the  power  for  the  whole  combination, 
w»_    ((    weight         "         "  " 

W  =    "         "       to  the  1st  in  the  series  and  power  to  the  2d, 
W"=    "         "  "      2d  "  "  "  3d, 

&c,  &c,  &c. 

Let  a1=  the  ratio  of  the  power  to  the  weight  in  the  1st, 
a2=  "  "  "  "  "     2d, 

a  3=  "  "  "  "  «      3d,&c. 

JL_     w'_     W"_  W"-'_ 

lhen  yy,  —  ai>  ^y//— a2'  W"/"- a3 ~yy^~~a" » 

and,  taking  the  product, 

P 
ypi=ax.a2.a3 aH. 

Hence  the  ratio  of  the  power  to  the  weight  in  any  combina- 
tion of  elementary  machines  is  equal  to  the  product  of  the  ratios 
of  the  power  to  the  weight  in  each  of  the  simple  machines. 

189.  Prop.  To  determine  the  ratio  of  the  power  to  the  pressure 
in  the  combination  of  levers  called  the  knee. 

This  combination  of  levers  is  used  with  advantage  where 
very  great  pressure  is  required  to  act  through  only  a  very 
small  space,  as  in  coining  money,  in  punching  holes  through 
thick  plates  of  iron,  in  the  printing-press,  &c      The  lever  AB 


BALANCES,    ETC. 


115 


2*-P 


turns  about  a  firmly  fixed  pivot 
at  A,  and  is  connected  by  anoth- 
er pivot  at  B  to  the  rod  BC, 
whose  extremity  C  produces  the 
pressure  on  the  obstacle  at  E. 

Let  the  power  P  act  horizon- 
tally at  some  point  F  in  the  le- 
ver AB,  ANC  be  a  vertical  line 
meeting  the  direction  of  P  in  N, 
and  DE  a  horizontal  plane,  on 
which,  at  E,  is  the  substance 
subject  to  pressure.  Let  R= 
the  reaction  of  the  rod  BC  in  the 
direction  of  its  length,  AM,  DL  perpendiculars  upon  its  direc- 
tion from  A  and  D,  and  W  the  vertical  resistance  of  the  sub- 
stance at  E. 

Taking  the  moments  about  A  and  D  in  equilibrium,  we  have 
P.AN=R.AM,  and  W.DE-R.DL, 
PAMxDE 
°r  W~ANxDL* 

When  BC  becomes  nearly  vertical,  DL  becomes  nearly  equal 

to  DE,  and  AN  to  AF,  while  AM  becomes  very  small. 

T     , .      .       .        P     AM  .  ,      _ . 

In  this  situation,  Tx7=~rFr  nearly,  so  that  P  is  a  very  small 

fraction  of  W. 


CHAPTER  VII. 


APPLICATION    OF     THE    PRINCIPLE    OF    VIRTUAL   VELOCITIES   TO    THJ 
MECHANICAL    POWERS. 

190.  In  Arts.  80  and  81,  it  is  shown  that  the  principle  of 
virtual  velocities  obtains  for  all  cases  of  equilibrium  of  a  free 
body  under  the  actions  of  any  number  of  external  forces  in  the 
same  plane. 

In  the  mechanical  powers,  the  parts  by  which  the  actions 
of  the  forces  are  transmitted  being  rigid  or  inextensible,  the 
forces  may  be  considered  as  acting  in  the  same  plane,  and  the 
internal  reactions  and  tensions  will  not  enter  the  fundamental 
equation  2.P.u=0.  Also,  the  virtual  velocities  of  the  support- 
ing parts  will  in  general  be  zero  for  the  possible  displacements 
of  the  system. 

In  some  of  the  mechanical  powers,  the  principle  applies  to 
all  possible  displacements,  however  great,  since  they  must  be 
in  the  direction  of  the  forces.  This  is  true  in  the  wheel  and 
axle,  toothed  wheels,  pulleys  with  parallel  cords,  the  inclined 
plane,  the  wedge,  and  screw.     In  the  lever,  and  pulleys  with 

inclined  cords,  the*  displace- 
ments must  be  taken  indefinitely 
small. 

191.  Prop.  The  principle  of 
virtual  velocities  obtains  in  the 
wheel  and  axle  in  equilibrium. 

The  forces  which  act  on  the 
wheel  and  axle  are  the  power 
P,  the  weight  W,  and  the  reac- 
tion R  of  the  steps  which  sup- 
port each  end,  C,  of  the  pivot 
about  which  it  turns,  and,  in  con- 
sequence of  the  rigidity  of  the 


PRINCIPLE     OF     VIRTUAL     VELOCITIES,     ETC.    117 


system,  they  may  be  considered  as  acting  in  the  same  plane. 
Also,  the  wheel  and  axle  receiving  a  displacement  turning 
about  C,  the  virtual  velocity  of  R  equals  0. 

Let  A  and  B  be  the  points  at  which  the  cords  left  the  wheel 
and  the  axle  respectively  before  displacement ;  A',  B'  after- 
ward. Then  W  ascends  through  the  space  WW-  arc  BB', 
and  P  descends  through  PP'=arc  A  A'.  PP'  is  the  virtual  ve- 
locity (Art.  78)  of  P,  and  positive  ;  WW  is  the  virtual  veloc- 
ity of  W,  and  negative.     Hence  (Art.  81), 

P.PP'-W.WW=0, 
or  P.arcAA'-W.arcBB'=0, 

or  P.AC.z.A'CA-W.BC.A.BCB'=0. 


P.AC-W.BC: 


W    AC 
°'  °r  P  =BC' 


the  condition  of  equilibrium  found  in  Art.  138. 

192.  Prop.   The  principle  of  virtual  velocities  obtains  in  a 
pair  of  toothed  wheels. 

Let  the  circles  in 
the  annexed  figure 
represent  the  pitch- 
lines  of  the  wheels 
(Art.  144),  andD1?D2 
the  points  which  were 
in  contact  in  the  line 
CC  before  displace- 
ment. Since  the  pitch- 
lines  roll  on  each  oth- 
er without  slipping, 
arc  DDj=arc  DD2, 
md 


P's  displacement  =PP=AC'.^D1C,D=AC 

W's  displacement  =WW  =CB.z.D2CD=CB 

By  the  principle  of  virtual  velocities, 
P.PP'-W.WW  =0, 


arcD.D 

CD    ' 

arc  D3D 

~CD~ 


ilS 


STATICS. 


or 


or 


RAC,i^-W.CB.^^=0, 


CD 


CD 


„AC    TirCB 

p w  ■ — =0 

DC         CD 

W    AC  CD  ■  W     CD 

;  and  when  AC  =  BC, 


"    P      BC'C'D 
as  found  in  Art.  143. 


P  ~'C'D' 


193.  Pkop.  TTie  principle  of  virtual  velocities  obtains  in  the 
single  movable  pulley  with  parallel  cords. 


If  the  pulley  A  be  raised  to  A', 
we  shall  have  AA'=WW'=iPF, 
since  each  of  the  cords  passing 
round  the  pulley  A  must  be  short- 
ened by  a  length  =WW.  WW, 
the  virtual  velocity  of  W,  is  nega- 
tive. 

.-.  P.PP-W.WW'=0, 
or  P.PP'-W.|PP'=0. 

W 

Hence  -p-= 2,  as  found  in  Art.  158. 


194.  Prop.  The  principle  of  virtual  velocities  obtains  in  the 
first  system  of  pulleys. 

In  the  figure  of  Art.  160  we  see  that,  if  W  be  raised  through 
a  space  WW',  each  of  the  n  cords  at  the  lower  block  will  be 
shortened  by  the  same  quantity,  or  that  P  will  descend  through 
a  space  n.WW.  Hence  the  equation  of  virtual  velocities 
P.PP-W.WW'=0  becomes 


P.7i.WW'-W.WW'=C  or  Y=n> 


as  in  Art.  160. 


PRINCIPLE     JF     VIRTUAL     VELOCITIES,    ETC.      119 

195.  Prop.  The  principle  cf  virtual  velocities  obtains  in  the 
second  system  of  pulleys. 

Referring  to  the  figure  in  Art.  161,  we  see  that,  if  P  de- 
scends through  the  space  PP', 

pp, 

the  pulley  an     would  rise  through  a  space  =_it~> 

PP' 


•i  > 


2 

&c,  &c, 

pp/ 

((  n  «  «  »  

"  3  2"-2' 

pp 

PP' 

"         a  j  or  the  weight  W  "  -r^-, 

And  the  equation  of  virtual  velocities  P.PP'— W.WW'=0 

becomes 

PP' 
P.PP'-W.— r=0, 

w 

or  -p-=2", 

the  same  as  in  Art.  161,  Cor.  2. 

196.  Prop.  The  principle  of  virtual  velocities  obtains  in  the 
third  system  of  pulleys. 

Referring  to  figure  of  Art.  162,  and  designating  the  pulleys 
as  in  that  article,  we  see  that,  if  W  be  raised  a  space  =WW, 
each  cord  will  be  shortened  by  a  space  equal  to  WW'.  The 
highest  movable  pulley  an_x  will  descend  a  distance  =WW. 
The  next  pulley  a„_2  will  descend  a  distance  =2. WW'  by  the 
descent  of  an_lf  and  a  distance  WW'  by  the  elevation  of  W,  oi 
will  descend  on  the  whole  (2+1) WW'. 

Similarly,  the  pulley  a„_3  will  descend  through 

{2(2  +  l)  +  l  SW.W'=(2'+2  +  l)WW'. 
Proceeding  in  the  same  way,  we  find  that  pulley  a„Hn_^)t  or 
a3,  will  descend  through  the  space 


120 


T  A  T  I  C  S. 


(2"-4+2"-*+,  &c 2+1)  WW, 

and  a0  through  the  space  (2"-3+2n-4+,&cc 2+1) WW, 

and  a",         "         «         "       (2"-2+2"-3+,&c 2  +  l)WW, 

and  P  will  descend  through  twice  the  last  found  space  by  the 
descent  of  the  pulley  al,  and  through  the  space  WW'  by  the 
elevation  of  the  weight ; 

or  PF=WW'.{2(2"-+2"-3+,  &c 2  +  l)  +  l} 

=  WW.(2"-1+2"-2+,  &c 22+2  +  l) 

=WW(2"-1). 
The  equation  of  virtual  velocities  is 

P.PP'-W.WW'=0, 
which  becomes 

P.WW(2"-1)-W.WW=0, 
W 


or 


as  in  Art.  162. 


=2"-l 


197.  Prop.  The  principle  of  virtual  velocities  obtains  in  the 
inclined  plane. 

Let  the  force  P  make  any  an- 
gle e  with  the  plane,  a=  Z-BAC, 
a  the  first  position  of  the  body 
whose  weight  is  W,  a'  the  posi- 
tion of  it  after  displacement. 

Drawing  the   perpendiculars 
av,  a'u,  we  have   —a'v=—aa' 
cos.  £=  the  virtual  velocity  of 
Yw'   V  w~  P,  and  au=aa'.  sin.  a—  the  vir- 

tual velocity  of  W. 

By  the  equation  of  virtual  velocities, 
¥.a'v-W.au=0, 
or  P.aa'.  cos.  e—W.aa'.  sin.  c=0. 

W     sin.  e 
Hence 

as  found  in  Art.  164. 


sin.  a 


198.  Prop.   The  principle  of  virtual  velocities  obtains  in  the 


wedge. 


PRINCIPLE     OF    VIRTUAL    VELOCITIES,    ETC.     121 


Let  2P  be  the  whole  power,  R 
and  R  the  pressures  perpendicu- 
lar to  the  faces  of  the  wedge  ABC, 
which  produce  equilibrium.  Let 
the  wedge  be  displaced  to  the  posi-  # 
tion  A'B'C.  The  displacement  of 
tne  point  of  application  of  Pis  aa'= 
AA' ;  that  of  b,  the  point  of  applica- 
tion of  R,  is  bb'=Am,  a  perpendicu- 
lar from  A  on  A'B',  and  Am=AA' 

.    BAC 

sin.  — - — . 


The  equation  of  virtual  velocities  is 
Y.aa'-R.bb'=0, 
BAC 


or 


P.AA'-R.AA'.  sin.  — — =0. 


R= 


BAC 


sin. 


as  found  in  Art.  169. 

199.  Prop.  The  principle  of  virtual  velocities  obtains  in  a  lev 
cr  of  any  form. 

Let  ACB  be  the  lever  be- 
fore displacement,  A'C'B' 
its  position  afterward. 
From  A'  draw  A'u  perpen- 
dicular to  AP,  and  from  B', 
B'w  perpendicular  to  BQ, 
produced.  Au  is  the  vir- 
tual velocity  of  P,  and  Bu 
that  of  Q.  Now  when  the  displacements  are  indefinitely 
small,  the  circular  arcs  AA',  BB'  become  straight  lines,  and 
kv=KA.i  cos.  A'Au-AC.Z.ACA'.  cos.  (PAC-900) 

= AC.  sin.  PAC.Z.ACA'; 
Bw=BC.  sin.  QBC.Z.BCB', 
and  aACA'  =  ^BCB'. 

The  equation  of  virtual  velocities  is 


122 


STATICS. 


~P.Av-Q,.Bu=0, 
or  P.AC,  sin  PAC-Q.BC.  sin.  QBC=0. 

_P_BC.  sin.  QBC 
*'•  Q~AC.sin.PAC 

as  found  in  Art.  134. 

200.  Prop.  The  principle  of  virtual  velocities  obtains  in  th& 
single  movable  pulley  with  cord's  inclined. 

Let  A  be  the  point  where 
the  cords  produced  would 
meet  at  the  first  position 
of  the  pulley,  when  P  and 
W  are  the  positions  of  the 
power  and  weight. 

Let  P  be  displaced  to 
P',  when  the  weight  is 
raised  to  W,  or  the  point 
f  of  meeting  of  the  cord  is 
raised  to  A'.  Draw  the 
circular  arcs  A'm,  A'n  with 
centers  B  and  C.  When  the  displacement  is  indefinitely  small, 
the  arcs  A'm,  A'n  become  straight  lines,  and 

Am=AA'  cos.  BAA'=Aw, 

BAC 
PP'=Am+A7i=2AA'  cos.-—-,  WW=AA*. 

The  equation  of  virtual  velocities  is 

P.PP'-W.WW'-O, 
which  becomes 

BAC 


P.2AA'.  cos. 
W 


or 


=2  cos. 


-W.AA' 
BAC 


0, 


P     2 

as  found  in  Art.  158. 

Scholium.  In  the  preceding  propositions,  the  expression 
P.PP'=W.WW, 


PRINCIPLE     OF    VIRTUAL    VELOCITIES,    ETC.     123 

WW'P 

or  PF  ~w 

explains  the  principle  that,  "  in  using  any  machine,  what  we 
gain  in  power  we  lose  in  time."  For,  in  order  that  W  may  be 
moved  through  any  given  space,  we  must  have  the  space 
moved  through  by  the  power  P,  increased  in  the  same  ratio 
that  P  is  diminished. 


CHAPTER   VIII. 

FRICTION. 

201.  The  surfaces  on  which  bodies  pressed  have  hitherto 
been  regarded  as  perfectly  smooth,  so  that  they  offered  no  re- 
sistance to  motion  parallel  to  themselves,  their  only  reaction 
being  perpendicular. 

When  rough  surfaces  are  in  contact,  the  motion,  or  tenden- 
cy to  motion,  parallel  to  the  surfaces,  is  affected  by  the  rough- 
ness, and  the  effect  is  called  Friction. 

Friction  may  be  divided  into  two  kinds :  sliding  friction, 
when  one  rough  surface  slides  on  another,  and  rolling  friction, 
when  one  rolls  on  the  other.  The  former  only  will  be  consid- 
3red  here,  under  the  term  Statical  Friction. 

202.  The  laws  of  friction  are  determined  by  experiment. 

.  If  the  body  A  rest  upon 

the     perfectly     smooth 
>-F   plane  BC,  the  smallest 


(27 


© 


35 


possible  force  applied  to 

it  will  cause  it  to  move. 
But  if  the  body  or  the 
plane,  or  both,  be  rough, 
a  force  within  certain 
limits  of  magnitude  may 
be  applied  to  it  without  causing  motion.  The  greatest  force 
which  can  be  so  applied  to  the  body  in  the  direction  of  the 
plane  will  measure  the  friction. 

Let  W  be  this  force,  acting  by  a  cord  over  a  pulley  on  the 
body  A,  F  being  the  opposing  force  of  friction.     Then  F=W. 
203.  The  following  laws  of  friction  are  deduced  by  this  or 
some  similar  process. 

1°.   The  friction  of  the  same  body,  or  a  body  of  the  same  ma- 
terial, when  the  weight  is  the  same,  is  independent  of  the  extent 


FRICTION. 


125 


of  the  surfaces  in  contact,  except  in  extreme  cases,  whete  the 
weight  is  very  great  compared  with  the  surfaces  in  contact. 
Thus  the  friction  of  the  body  A  will  be  the  same  whichever 
side  rest  on  the  plane,  or  whatever  be  the  form  within  the  ex- 
cepted limits. 

2°.  The  friction  is  proportional  to  the  pressure  on  the  plane, 
or  the  reaction  of  the  plane,  within  moderate  limits.  If  other 
weights,  as  m,  be  placed  on  A,  W  or  F  will  vary  as  the  whole 
reaction  R  of  the  plane. 

Cor.  The  friction  is  therefore  some  function  of  this  pressure, 
and  we  may  represent  it  by  juR.     ju.  is  called  the  coefficient  of 

F 

friction,  and  is  equal  to  ^5-,  or  the  ratio  of  the  friction  to  the  re- 
action of  the  plane. 

204.  Prop.  The  coefficient  of  friction  between  two  given  sub- 
stances is  equal  to  the  tangent  of  the  inclination  of  the  plane 
formed  of  one  of  the  substances,  when  the  body  formed  of  the  oth- 
er is  about  to  slide  down  it. 

Let  the  inclination  a  of  the  plane  AC  be  increased  till  the 
body  a  is  just  on  the  point 
of  sliding  down  it.  The  body 
a  will  then  be  in  equilibrium 
from  the  normal  reaction  of 
the  plane  R,  the  friction  juR 
acting  up  the  plane,  and  its 
weight  W  acting  vertically 
downward. 

Resolving  parallel  and  perpendicular  to  the  plane,  we  have 

[iR  —  W.  sin.  a=0, 
R— W  cos.  a=0. 

Eliminating  R,  we  have 


or 


ft  cos.  a  — sin.  a=0, 
ju=tan.  a. 


205.  Prop.   To  determine  the  limits  of  the  ratio  of  P  to  W  on 
an  inclined  plane,  when  friction  acts  up  or  down  the  plane. 


126 


STATICS. 


Let  the  power  P  make   an 
angle  e  with  the  plane  whose 
inclination    is    a,    and    W    the. 
weight  of  the  body. 

1°.  To  determine  the  great- 
est value  P  can  have  without 
causing  the  body  to  move  up 
the  plane.  In  this  case,  the 
friction  juR,  opposing  the  motion 
up  the  plane,  will  act  down  it,  and,  resolving  parallel  and  per 
pendicular  to  the  plane,  we  have 

P.  cos.  e—fiR—W  sin.  a—0,  (a) 

P.  sin.  e+  R-Wcos.  a=0.  (b) 

Multiplying  (b)  by  /z  and  adding  to  (a),  we  have 
P(cos.  e+n  sin.s)  — W.  (sin.  a-\-\i  cos.  a)  —  0, 

_,     W.  (sin.  a -\-fxcos.  a) 
or  P= * : : . 

COS.  £+jU  sin.  £ 

2°.  To  determine  the  least  value  P  can  have  without  causing 
the  body  to  move  down  the  plane.  In  this  case  the  friction 
will  oppose  the  descent,  and  will  therefore  act  up  the  plane. 
Hence  equations  (a)  and  (b)  become 

P.  cos.  e+j"R— W  sin.  <z=0,  (a') 

P.  sin.  e+  R- W  cos.  a=0.  (V) 

Multiplying  (b1)  by  ^  and  subtracting,  we  find 
W.  (sin.  a— \i  cos.  a) 
cos.  e—fj,  sin.  e 
No  motion  will  take  place  while  the  value  of  P  is  between 
these  two,  which  are  its  limits. 

These  two  values  of  P  may  be  combined  so  as  to  take  the 
form 

W     cos.  e±jj,  sin.  e 
P      sin.  a±ju  cos.  a 
in  which  the  upper  sign  is  to  be  taken  when  friction  acts  down 
the  plane,  and  the  lower  when  the  friction  acts  up  the  plane. 

206.  Prop.  To  determine  the  limits  of  the  ratio  of  P  to  W  in 
the  screw,  when  friction  acts  assisting  the  power  or  the  weight. 


FRICTION. 


127 


Proceeding  as  \nArt. 
173,  let  ABC  be  the 
inclined  plane  formed 
by  unwrapping  one 
revolution  of  the 
thread,  the  angle  BAC 
=a  ;  let  W=  the 
whole  weight  sustain- 
ed by  the  screw,  w= 
that  part  of  it  support- 
ed at  a,  Q=  the  whole 
force  acting  at  the  cir- 
cumference of  the  cyl- 
inder, r= ED  the  radius  of  the  cylinder,  a—FD  the  lever  at 
which  the  power  acts,  and  q  that  part  of  Q,  which  supports  w 

at  a.     Then  Q=P-. 
r 

The  forces  which  are  in  equilibrium  at  a  are  the  weight  w, 
the  reaction  R,  the  horizontal  force  q,  and  the  friction  j^R  act- 
ing up  or  down  the  plane. 

Resolving  parallel  and  perpendicular  to  the  plane,  we  have 
q.  cos.  a±juR  —  w.  sin.  a=0,  (a) 

q  sin.  a—  R+w  cos.  a=0.  (b) 

Multiplying  (b)  by  jw,  and  adding  and  subtracting,  we  have 
<7(cos.  a±ft  sin.  a)  —  w(s'm.  azjp\i  cos.  a)=0, 
q  _  Q,  _sin.  a^zfi  cos.  a 
w 
JP 

•'•  W 


or 


W    cos.  a±//  sin.  a 
r  sin.  a=pj«  cos.  a 


a  cos.  a±jit  sin.  a 
The  two  values  of  this  expression  give  the  limits  required, 

P 

and  ^tj  may  have  any  intermediate  value. 

207.  Cor.  From  the  two  preceding  propositions  it  will  ap- 
pear that,  when  we  have  obtained  one  of  the  limits,  the  other 
may  be  had  by  simply  changing  the  sign  of  fi  in  the  former. 


128 


STATICS. 


EXAMPLES. 


Ex.  1.  A  uniform  straight  beam  rests  on  a  rough  cylinder 
of  given  radius  ;  required  the  greatest  weight  that  can  be  sus- 
pended from  one  end  of  the  beam  without  causing  it  to  slide  off. 

Let  g  be  the  center  of  gravity  of  the 
beam  BC,  whose  length  is  2d,  and  Bg 
—  d  since  the  beam  is  uniform,  w=  its 
\v        9j(\  weight,  and  W  the  weight  suspended 

\A/^-"~-^\  from  B.      Before  the  weight  W  was 

suspended  from  the  beam  the  point  g 
must  have  been  at  A.  Let  A'  be  the 
point  of  contact  with  the  cylinder  when 
the  beam  is  on  the  point  of  sliding  off, 
a  the  angle  it  then  makes  with  the  ho- 
rizon, and  r—  the  radius  of  the  cylinder. 
Resolving  parallel  and  perpendicular  to  the  beam,  we  have 


juR— w.  sin.  a— W  sin.  a=0, 
R  —  w  cos.  a— W  cos.  a=0  ; 

whence  ju=  tan.  a. 

Taking  the  moments  about  A',  we  have 

W.BA'  cos.  a—iv.A'g.  cos.  a=0, 
or  W.(Bg—A'g)-iu.A'g=0. 

But  A'g=  arc  A'A=  radius  X  angle  AOA'=r.a. 

.-.  W.(d—  ra)—  w.ra=0, 


(a) 


or 


W= 


w.ra      w.rtan.  lfi 


,  the  weight  required. 


d—ra     d—r tan.  xfi 

Ex.  2.  A  ladder  rests  with  one  end  on  a  rough  horizontal 
plane,  and  the  other  on  a  rough  vertical  wall ;  given  1=  its 
length,  d—  the  distance  of  its  center  of  gravity  from  its  lower 
end,  (i  and  [j,'=  the  coefficients  of  friction  on  the  horizontal  and 
vertical  planes  respectively;  required  its  inclination  6  when 
on  the  point  of  sliding  down. 

Let  AB  be  the  ladder  and  the  forces  acting  upon  it,  as  rep- 
resented in  the  figure. 


EXAMPLES     ON     CHAPTERS     V.       VII.,    AND     VIII.    129 

Resolving     horizontally     and 
vertically,  we  have 

R'_^R  =  0,  (a) 

n+fj,'R'-w=o.    (b) 

Multiplying  (a)  by  y,',  and  sub- 
tracting from  (b),  we  find 

w  =  R(l+W'), 
and  from  (a),  R'=[j,R. 

Taking  moments  about  A,  we 
get 

w.d.  cos.  e—R'.l  sin.  d—n'R'J.  cos.  6=0, 
w.d-fi'R'l 


or 


tan.  6- 


R'l 


Substituting  the  values  of  w  and  R', 

d.(l+nn')  —  1\l\l' 


tan.0=- 


lil 


If  the  center  of  gravity  of  the  ladder  be  at  the  middle  point, 
l=2d,  and 

1—  n/tf 


tan.  6=- 


2[i 


208.    EXAMPLES    ON    CHAPTERS    VI.,  VII.,  AND    VIII. 

Ex.  1.  A  beam  30  feet  long  balances  on  a  prop  %  of  its 
length  from  the  thicker  end;  but  when  a  weight  of  10  lbs.  is 
suspended  from  the  other  end,  the  prop  must  be  moved  2  feet 
toward  it  to  maintain  the  equilibrium.  Required  the  weight 
of  the  beam. 

Ex.  2.  The  forces  P  and  Q,  act  at  arms  a  and  b  respectively 
of  a  straight  lever,  which  rests  on  a  fixed  point  to  which  it  is 
not  attached.  When  P  and  Q,  make  angles  a  and  (3  with  the 
lever,  required  the  conditions  of  equilibrium. 

Ex.  3.  A  uniform  beam  is  sustained  by  three  persons,  one 
at  one  end,  and  the  other  two  by  a  hand-spike  placed  at  some 

I 


M\, 


130  STATICS. 

point  beneath  it.  At  what  point  must  the  hand-spike  be  placed 
that  each  person  may  sustain  one  third  of  the  weight  ? 

Ex.  4.  A  Roman  steelyard,  whose  weight  is  10  lbs.,  has  its 
center  of  gravity  2  inches  from  the  fulcrum,  and  the  weight  to 
be  determined  is  supported  by  a  pan  placed  at  a  distance  of  3 
inches  on  the  other  side.  Find  the  respective  distances  from 
the  fulcrum  at  which  the  constant  weight  of  5  lbs.  must  be 
placed,  in  order  to  balance  10,  20,  30,  &c.,lbs.  placed  success- 
ively in  the  pan. 

Ex.  5.  Find  the  ratio  of  the  power  to  the  pressure  in  the 
common  vice. 

Ex.  6.  Find  the  ratio  of  the  power  to  the  pressure  in  the 
screw  by  the  principle  of  virtual  velocities. 

Ex.  7.  An  isosceles  triangle,  whose  base  is  to  one  of  its 
equal  sides  as  1  :  Vl,  is  placed  with  its  base  on  an  inclined 
plane ;  and  it  is  found  that,  when  the  body  begins  to  slide,  it 
also  begins  to  roll  over.     Find  the  coefficient  of  friction. 

Ex.  8.  A  ladder  rests  against  a  vertical  wall,  to  which  it  is 
inclined  at  an  angle  of  45°  ;  the  coefficients  of  friction  of  the 
wall  and  of  the  horizontal  plane  being  respectively  \  and  \, 
and  the  center  of  gravity  of  the  ladder  being  at  its  middle 
round.  A  man  whose  weight  is  half  the  weight  of  the  ladder 
ascends  it.  Find  to  what  height  he  will  go  before  the  ladder 
begins  to  slide. 

Ex.  9.  In  a  uniform  lever  of  the  second  kind,  which  weighs 
2  ounces  per  inch,  required  the  length  of  the  lever,  in  order 
that  the  power  may  be  the  least  possible  when  in  equilibrium 
with  a  weight  of  48  ounces  placed  at  a  distance  of  three  inches 
from  the  fulcrum. 

Ex.  10.  Two  given  weights,  P  and  Q,  are  suspended  from 
two  given  points  in  the  circumference  of  the  wheel,  a  being 
the  angle  made  by  the  radii  drawn  to  the  points  of  suspension. 
Required  the  angle  6  which  the  lower  radius  makes  with  the 
vertical  when  the  weights  cause  the  greatest  pressure  on  the 
axle. 


**/ 


B 


V 

f                        A        %" 

e 

1 

A 

\ 

1 
4* 

8 x    f° 


(3  dc^r  a  -    T 


MP 


DYNAMICS. 


INTRODUCTION.— DEFINITIONS. 

209.  In  Statics  we  have  investigated  the  relations  of  the  in- 
tensities and  directions  of  forces  necessary  to  produce  equilib- 
rium, this  result  being  entirely  independent  of  the  time  during 
which  the  forces  act. 

In  Dynamics,  forces  are  regarded  as  producing  motion  or 
change  of  motion  in  bodies,  and  these  effects  must  obviously 
depend  on  the  duration  of  the  action  of  the  forces.  In  dy- 
namics, therefore,  time  becomes  an  element  in  our  investiga- 
tions. 

210.  Motion  is  the  transit  of  a  material  point  or  body  from 
one  position  to  another  in  space. 

211.  The  absolute  motion  of  a  body  is  its  transit  from  one 
fixed  point  in  space  to  another. 

212.  Relative  motion  is  a  change  of  distance  from  a  point 
which  is  itself  in  motion. 

All  motions  are  relative  in  any  practical  view  which  we 
can  take  of  them,  since  we  have  no  means  of  determining  the 
absolute  rest  of  any  point  in  space. 

213.  The  velocity  of  a  body  is  its  rate  of  motion.  It  may  be 
uniform  or  variable. 

The  velocity  of  a  body  is  uniform  when  it  passes  over  equal 
spaces  in  equal  times,  and  is  measured  by  the  space  passed  over 
in  the  unit  of  time. 

Let  v  be  the  velocity,  or  space  passed  over  in  one  second, 
then  the  space  described  in  two  seconds  will  be  2v,  in  three 
seconds  3u,  and  so  on ;  and  if  s  be  the  space  described  from 
the  commencement  of  motion,  and  t  the  number  of  seconds 
also  reckoned  from  the  commencement  of  motion,  then 


132  DYNAMICS. 

s=vt.  [I.] 

The  units  of  time  and  space  are  arbitrary.  It  is  usual  to  take 
one  second  for  the  unit  of  time,  and  one  foot  for  the  unit  of 
space.  When  no  mention  is  made  of  different  units,  these  will 
be  understood. 

214.  Variable  velocity  is  that  which  continually  increases  or 
decreases,  so  as  to  be  the  same  at  no  two  successive  instants. 
To  find  a  measure  of  the  velocity  of  a  point  or  body  so  moving, 
let  us  assume,  at  first,  that  the  velocity  which  the  body  has  at 
the  end  of  t1  seconds  is  uniform  from  t1  seconds  to  t2  seconds, 
a  very  small  interval,  the  space  passed  over  at  the  end  of  tt 
seconds  being  slf  and  the  space  passed  over  at  the  end  of  ta 
seconds  being  s2.  Then,  using  the  symbol  A  to  signify  finite 
difference,  by  [I.]  we  have 

s2—s.     As 

w=— l=T~- 

t2-t1     A* 

If,  however,  there  is  no  time,  however  small,  during  which  the 
velocity  is  uniform,  then  the  smaller  we  take  &t,  and  conse- 
quently As,  the  more  nearly 

That  is,  the  velocity,  when  variable,  is  measured  by  the  limit 
of  the  ratio  of  the  space  described  to  the  time  of  describing  it. 

215.  Relative  velocity  is  the  velocity  with  which  two  bodies 
approach  or  recede  from  each  other. 

216.  Matter  at  any  given  moment  must  be  in  one  of  the  two 
states,  motion  or  rest.  The  inertia  of  matter  is  the  entire  ab- 
sence of  power  in  itself  to  change  this  state.  It  implies  equal- 
ly a  disability,  when  in  motion,  to  change  its  rate  or  its  direc- 
tion.    Hence 

A  body,  when  not  acted  on  by  any  external  forces,  if  at  rest, 
will  remain  so,  or,  if  in  motion,  will  continue  to  move  in  a,  straight 
line  and  with  a  uniform  velocity. 

This  is  called  the  first  law  of  motion. 

217.  It  is  a  consequence  of  the  inertia  of  matter,  that  when 
a  force  is  applied  to  a  body  to  move  it,  each  of  its  particles 
opposes  a  resistance  to  motion  in  directions  parallel  but  oppo- 


DEFINITIONS.  133 

site  to  the  direction  of  the  applied  force.  The  center  of  these 
parallel  forces  (Art.  44)  of  resistance  is  called  the  center  of  in- 
ertia of  the  body.  It  is  the  same  point  which,  in  statics,  was 
called  the  center  of  gravity  of  the  body  in  reference  to  the 
force  which  was  there  supposed  to  act  on  the  body. 

The  sum  of  these  parallel  resistances,  or  their  resultant 
(Art.  43),  is  obviously  proportional  to  the  number  of  particles 
in  the  body  or  to  the  whole  mass.  Hence  the  inertia  of  a 
body  is  a  surer  test  of  the  quantity  of  matter  or  mass  of  a 
body  than  its  weight  is ;  for  the  latter  (Arts.  84  and  88)  varies 
by  a  change  of  position  on  the  earth,  while  the  former  is  al- 
ways the  same. 

218.  The  path  of  a  body  is  the  line,  straight  or  curved, 
which  its  center  of  inertia  describes  when  it  passes  from  one 
point  to  another  in  space. 

219.  A  body  is  said  to  he  free  or  move  freely  when  its  path 
depends  on  the  action  of  the  impressed  forces  only.  Its  mo- 
tion is  said  to  be  constrained  when  its  path  is  limited  to  a  given 
line,  straight  or  curved,  or  limited  to  a  given  surface. 

220.  An  impulsive  force  is  one  which  acts  instantaneously 
and  without  sensible  duration. 

221.  An  incessant  force  is  one  which  acts  without  intermis- 
sion. If  a  material  point  move  from  rest  by  the  action  of  an 
incessant  force,  its  rate  of  motion  or  velocity  must  continually 
increase.  The  amount  of  this  increase,  or  the  increment  of  the 
velocity  in  the  unit  of  time,  will  obviously  be  greater  or  less  as 
the  intensity  of  the  force  is  greater  or  less.  This  increment 
of  the  velocity  in  one  second  is  therefore  the  measure  of  the 
intensity  of  the  force. 

222.  A  constant  force  is  an  incessant  force  whose  intensity 
is  at  all  times  the  same.  If  a  material  point  move  by  the  ac- 
tion of  a  constant  force,  the  increments  of  the  velocity  in  each 
successive  unit  of  time  must  all  be  equal,  and  each  increment 
will  be  a  measure  of  the  force.  If,  therefore,  we  put  4>  for  the 
increment  of  velocity,  or  the  velocity  generated  by  the  force 
in  one  second,  <p  will  represent  the  force.  The  increments  be- 
ing all  equal,  the  velocity  generated  in  two  seconds  will  he 


134  DYNAMICS. 

20,  in  three  seconds  30,  and  so  on.  Hence,  if  v  be  the  velocity 
generated  in  t  seconds, 

v=<pt.  [III.] 

223.  A  variable  force  is  an  incessant  force  whose  intensity 
either  increases  or  decreases,  so  as  to  be  the  same  at  no  two 
successive  instants.  To  find  an  expression  for  a  variable 
force,  let  us  assume  it  to  be  constant  from  the  end  of  the  time 
t ,,  when  the  velocity  is  d,,  to  the  end  of  the  time  t2,  when  the 
velocity  is  v2.     Then,  by  [III.],  we  have 

v2—vl     Au 

ta—tl  ~a7* 

If,  however,  there  be  no  interval  during  which  the  force  is 
constant,  then  the  smaller  we  take  A/,  and  consequently  At> 
the  more  nearly  will 

Au 

That  is,  a  variable  force  is  measured  by  the  limit  of  the  ratio 
of  the  velocity  caused  by  it  to  the  time  of  causing  it. 

224.  The  momentum  of  a  body  is  its  quantity  of  motion,  ana 
is  measured  by  the  product  of  the  mass  of  the  body  by  its  veloc- 
ity. For  the  motion  of  a  single  particle  is  its  velocity,  and  the 
motion  of  any  number  of  particles,  having  the  same  velocity, 
is  obviously  as  much  greater  as  the  number  of  particles  is 
greater.  Hence  the  whole  motion  is  equal  to  the  whole  num- 
ber of  particles  in  the  body,  or  its  mass,  multiplied  by  their 
common  velocity'.  Or,  if  Q,  be  the  quantity  of  motion  of  a 
body,  M  its  mass,  and  V  its  velocity, 

Q=MV.  [V.] 

The  mass,  multiplied  by  the  square  of  the  velocity,  is  called  the 
vis  viva,  or  the  living  force  of  a  body. 

225.  In  estimating  the  effects  of  incessant  forces,  we  have 
considered  only  the  acceleration  or  velocity  which  each  force 
will  produce  when  acting  on  a  free  material  point  or  a  unit  of 
mass.  When  so  measured,  they  are  called  accelerating  forces. 
If  the  mass  moved  differs  from  that  which  we  have  called  the 
unit  of  mass,  and  it  is  taken  into  consideration  in  estimating 


DEFINITIONS.  135 

the  effects  of  the  forces,  they  are  then  called  moving  forces. 
Let  0  be  the  acceleration,  or  velocity  generated  in  a  unit  of 
time,  M  the  number  of  units  of  mass  ;  then  $,  the  moving 
force,  will  be  measured  by  the  quantity  of  motion  generated 
in  the  unit  of  time,  or 

$=0M.  [VI.] 

$ 
Hence  0=tT|»  or  the  accelerating  force,  is  equal  to  the  moving 

force  divided  by  the  mass. 

226.  If  a  body  already  in  motion  be  acted  on  by  a  force  in 
the  direction  in  which  the  body  moves,  the  superadded  motion 
is  just  the  same  as  that  which  would  have  been  produced  in 
the  body  if  at  rest  when  the  force  began  to  act. 

If  the  force  act  in  a  direction  different  from  that  in  which 
the  body  moves,  the  new  motion  produced  by  the  force,  esti- 
mated in  the  direction  of  the  force,  will  be  the  same  as  if  the 
body  had  been  at  rest. 

If  the  force  act  in  a  direction  opposite  to  that  in  which  the 
body  moves,  the  motion  destroyed  in  the  body  is  equal  to  that 
which  the  force  would  produce  in  the  body  if  at  rest  when  the. 
force  began  to  act. 

These  facts  are  consequences  of  the  inertia  of  matter,  and 
will  receive  additional  illustration  in  the  sequel.  They  are 
embodied  in  the  following  enunciation,  called  the  second  law 
of  motion: 

All  motion  or  change  of  motion  in  a  body  is  proportional  to 
the  force  impressed  and  in  the  direction  of  that  force. 

227.  When  one  body  impinges  on  another  at  rest  or  in  mo- 
tion, the  quantity  of  motion,  or  momentum  of  the  two  bodies 
after  impact,  is  the  same  as  before  impact.  For  matter  being 
incapable  of  originating  motion,  can  not  add  to  the  motion  of 
other  matter,  or  take  from  it  except  by  imparting  its  own  mo- 
tion. Hence  whatever  motion  the  second  body  receives  in 
the  direction  of  the  striking  body,  just  so  much  must  be  lost  by 
the  striking  body. 

This  fact  is  usually  enunciated  as  follows,  and  is  called  the 
third  law  of  motion  : 

Action  and  reaction  are  equal  and  in  oppositt  directions. 


CHAPTER   f. 

UNIFORM     MOTION. 

228.  In  considering  the  effect  of  an  impulsive  force,  we  shall 
suppose  the  force  applied,  at  the  center  of  inertia  (Art.  217). 
When  so  applied  the  parallel  forces  of  resistance  of  all  the 
particles  situated  on  opposite  sides  of  the  direction  of  the 
force  will  balance  each  other,  and  the  body  will  not  rotate, 
but  all  its  particles  will  describe  parallel  lines  with  a  common 
velocity. 

229.  Prop.   To  find  the  general  equation  of  uniform  motion. 
X .        Let  OB  be  the  path  of  the  body  (Art.  218),  v  its  veloci- 

;  ty  (Art.  213),  and  t  the  time  of  its  motion  in  seconds. 
0  :  Let  O  be  the  origin,  or  point  from  which  we  estimate  the 
^ :  successive  positions  of  the  body,  s  the  distance  of  the  body 
from  O  at  the  end  of  the  time  t,  and  OA=Sj  its  distance 
from  the  origin  O  at  the  commencement  of  the  time. 
When  the  body  is  at  B,  we  have  [I.] 

AB=vt. 
But  OB=OA+AB, 

B      or  s=s,+vt,  (36) 

which  is  the  general  equation  of  uniform  motion. 

If  the  body,  instead  of  receding  from  the  origin  O,  approach 
it,  then  v  will  be  negative ;  or  if  the  body  begin  to  move  from 
A',  then  sz  will  be  negative. 

Cor.  1.  From  (36)  we  obtain 

s— s,     AB 

v=-r=— ■ 

or  the  velocity  of  a  body  is  equal  to  the  constant  ratio  of  the 
space  described  to  the  time  of  describing  it. 

Cor.  2.  If  we  estimate  the  position  of  the  body  from  the 


UNIFORM    MOTION.  137 

point  where  the  body  is  when  t=0,  or  suppose  the  space  and 
time  to  commence  together,  then  s1=0,  and 

s=vt.  (37) 

230.  Prop.  If  two  bodies  move  during  the  same  time,  their 
velocities  will  be  proportional  to  the  spaces  described  by  them 
respectively. 

Taking  the  points  of  departure  for  the  origin  of  spaces,  we 
have  from  (37) 

s=vt  and  s'=v't'. 
.'.  s  :  s'=vt  :  v't'. 
And,  since  t=t',  s  :  s'=v  :  v'. 

231.  Prop.  If  the  velocities  of  two  bodies  are  equal,  the  spaces 
described  are  proportional  to  the  times. 

As  before,  s  :  s'=vt  :  v't' ; 

and  if  u— v',  s  :  s'=  t  :  V. 

232.  Prop.  If  the  spaces  described  by  two  bodies  are  equal, 
their  velocities  are  reciprocally  proportional  to  the  times. 

For,  since  s  :  s'—vt  :  v't', 

ifs=s',  vt=v't'. 

.'.  v  :  v'=t'  :  t. 

233.  Prop.  An  impulsive  force  is  measured  by  the  momentum 
/it  can  produce  in  any  mass. 

If  v  be  the  velocity  produced  by  the  force  F,  in  a  body  con- 
taining M  units  of  mass,  by  [V.]  the  momentum  will  be  Mr. 
Now  if  the  mass  M  move  from  rest  by  the  action  of  the  force, 
all  the  motion  it  receives  is  the  effect  of  the  force.  And,  ad- 
mitting the  principle  that  effects  are  proportional  to  their 
causes,  if  A  be  the  constant  ratio  of  the  force  to  the  momentum, 
then 

F1=A.ilftj. 
But  the  unit  of  force  being  arbitrary,  we  may  assume  it  to  be 
A.     Hence,  putting  F  for  the  number  of  units  of  force,  or  the 
ratio  of  F,  to  A,  we  have 

F 
F=-^  =  Mu.  (38) 

Cor.  If  M=l,  F,=A.u,  in  which  X  is  constant.  Hence  the 
force  is  proportional  to  the  velocity  produced  in  the  unit  of  mass. 


138 


DYNAMICS. 


Schol.  Since  we  know  nothing  of  the  nature  of  forces,  we 
can  not  determine  their  effects  a  priori.  The  foregoing  propo- 
sition ought,  therefore,  to  be  regarded  as  depending  ultimately 
on  observation  and  experiment.  The  fact  that  on  the  earth, 
which  is  subject  to  the  double  motion  of  rotation  and  transla- 
tion, forces  are  found  to  produce  precisely  the  same  effects  as 
if  the  earth  were  at  rest,  is  a  confirmation  of  its  truth ;  also 
that  a  pendulum  performs  its  vibrations  in  the  same  time,  what- 
ever be  the  direction  of  these  vibrations  in  reference  to  an  east 
and  west  line. 

234.  If  any  number  of  forces  act  upon  a  body  in  the  same 
direction,  the  velocity  imparted  to  the  body  will  be  equal  to  the 
sum  of  the  velocities  imparted  by  each. 

Let  F,  F',  F"  .  .  .  be  any  number  of  forces  acting  upon  a 
body,  v,  »>',  v"  .  .  .  the  velocities  imparted  by  each  respective- 
ly, (j)  the  resultant  force,  and  u  the  velocity  due  to  <p. 

Then  F=Au,  F'=Au',  F"=Ai>",  &c.  Since  the  forces  are 
conspiring,  their  resultant  will  be  equal  to  their  sum. 

.-.  F+F'+F"+  &c,  =cp=X(v+v'+v"+  &c). 
But  (p  —  Xu. 
Hence  u=v-\-v'\-v"-±-  &c. 

235.  Prop.  If  the  adjacent  sides  of  a  parallelogram  represent 
in  magnitude  and  direction  the  velocities  which  two  forces,  by 
their  separate  action,  would  respectively  produce  in  a  body,  the. 
diagonal  of  the  parallelogram  will  represent  the  actual  velocity 
produced  by  their  joint  action. 

Let  AB  and  AC  represent  the  velocities  which  the  forces 

P  and  Q  respectively  im- 
part to  a  body  placed  at 
A,  or,  in  other  words,  the 
spaces  over  which  the 
body  would  pass  in  a  unit 
of  time  by  the  separate  ac- 
tion of  the  forces,  then  AD 
will  represent  the  velocity 
produced  by  their  result- 
ant,   or    the    space    ovei 


UNIFORM     MOTION. 


139 


which  the  body  will  pass  in  the  unit  of  time  by  the  joint  action 
of  the  forces. 

We  may  assume  AB  to  represent  the  force  P  ;  then,  by  Art. 
233,  Cor., 

P  :  Q=AB  :  AC, 
and  AC  will  represent  the  force  Q.     Also,  'Art*  21,  AD  will 
be  the  resultant  R  of  P  and  Q.     Let  x  be  the  velocity  due  to 
the  resultant  R,  then  (Art.  233,  Cor.) 
P  :  R=AB  :  x. 
But  P:R=AB:AD. 

.-.  s=AD, 
and  AD  being  the  direction  of  the  resultant,  the  velocity  due 
to  it  will  have  the  same  direction. 

The  two  preceding  propositions  illustrate  the  second  law 
of  motion  (Art.  226). 

236.  Prop.  The  velocity  of  a  body  being  given,  to  find  the  com- 
ponent velocities  in  any  directions  at  right  angles  to  each  other ; 
and  two  component  velocities  at  right  angles  to  each  other  being 
given,  to  find  the  resultant  velocity. 


Let  AD  represent  the  velocity  v  of 
the  body  in  direction  and  magnitude, 
AX  and  AY  the  rectangular  directions 
in  which  the  components  are  required, 
and  6  the  angle  which  AD  makes  with 
AX.  Completing  the  rectangle,  AB 
will  represent  the  velocity  a  in  AX, 
and  AC  the  velocity  b  in  AY. 


Then  AB=AD.  cos.  0,  AC=AD.  sin.  0, 

or  a=v.  cos.  9,  b=v.s'm.O.  (39) 

Again,  if  a  and  b  represent  the  given  component  velocities 
in  AX  and  AY,  to  find  the  resultant  velocity  v  we  have 

v=  Va'+b%  (40) 

b 

(41) 


and 

as  required. 


tan.0=-, 
a 


l40  DYNAMICS. 

Cor.  Hence  velocities  may  be  compounded  and  resolved 
like  forces  in  statics. 

237.  Prop.  Two  bodies,  A  and  B,  describe  the  same  path  with 
the  velocities  v  and  v',  and  at  the  commencement  of  their  motion 
are  at  a  distance  a  from  each  other :  to  find  the  time  t  when  they 
will  be  at  a  distance  b  from  each  other  and  the  position  of  each 
at  the  end  of  that  time. 

Taking  the  position  of  A  when  t=0  for  the  origin  of  spaces, 

the  equations  of  their  motions  are,  by  (37)  and  (36), 

s=vt,  (a) 

and  s'=a-{-v't'.  (b) 

By  the  conditions  of  the  question,  t=t'. 

Also  s—s'=b  or  s'—s=b,  .'.  s— s'=±6. 

Subtracting  (b)  from  (a),  and  putting  t=t',  we  have 

s— s'=vt— a— v't=±b. 

a±b  ,     x 

.-.  t= -.  (42) 

v — v 

a±b 
Also,  from  (a)  s=vt=v. ;,  (43) 

.  .         ,_.  a±b     va±v'b  ,     . 

and  irom  (b)        s'—a+v't—a+v'. -= — .  (44) 

»  '  v  —  v        V  —  V 

Cor.  1.  If  the  bodies  move  in  opposite  directions,  v'  will  be 
negative,  and 

a±b  a±&  va^iv'b 

v+v  v-\-v  v+v 

Cor.  2.  If  the  time  when  the  bodies  are  together  be  required, 
then  6=0,  and 

a  va  v'a 

t= .-,  s= ,=s',  s'  —  a= -. 

v — v  v — v  v — v 

238.  Prop.  When  two  bodies  A  and  B  move  in  the  circum 
ference  of  a  circle  with  uniform  velocities,  to  determine  the  cir- 
cumstances of  their  motion.  ■ 

Let  v  and  v'  be  their  velocities,  c  the  circumference  of  the 
circle,  and  a  the  distance  apart  at  the  commencement  of  the 
time.     Then,  putting  6=0  in  (42),  we  have 


UNIFORM     MOTION.  141 


t.  = =  the  time  of  their  first  meeting. 

v — v 

a+c 

ta= 7  =        "  "       second     » 

a+2c  ,.   , 

*Q  = r  =        "  "       third        " 


'3 


v— V 


a+(n— l)c 

tn= ^ — -!-=        "  "        nth 

v~v' 

also,  s=v — ; —  =  space  described  by  A, 

v— v 

.  v'a+v(n—l)c  _ 

and       s-a= —, — —=       "  "  B. 

v—v' 

Cor.  The  interval  between  two  successive  conjunctions  is 


239.    EXAMPLES. 

Ex.  1.  If  an  iron  rod  have  one  end  against  the  sun  and  the 
other  resting  on  the  earth,  the  distance  of  the  sun  from  the 
earth  being  95,125,000  miles,  in  what  time  will  a  blow  applied 
to  the  end  on  the  earth  be  felt  by  the  sun,  the  velocity  of  an 
impulse  in  iron  being  11,865  feet  per  second  ? 

Ans.  490  days. 

Ex.  2.  When  the  earth  is  in  that  part  of  its  orbit  nearest  to 
Jupiter,  an  eclipse  of  one  of  Jupiter's  satellites  is  seen  1G  min- 
utes 36  seconds  sooner  than  it  would  be  if  the  earth  were  in 
that  part  of  its  orbit  most  remote  from  Jupiter.  The  radius 
of  the  earth's  orbit  being  95,125,000  miles,  what  is  the  veloci- 
ty of  light? 

.  Ex.  3.  The  star  61  Cygni  is  ascertained  to  be  56,319,996,- 
600,000  miles  distant  from  us ;  were  its  light  suddenly  extin- 
guished, in  what  time  would  the  intelligence  reach  us,  the 
velocity  of  light  being  191,000  miles? 

Ex.  4.  Suppose  964  tons  of  ice  to  be  floating  directly  to  the 
east  at  sunrise  on  the  21st  of  March,  with  a  velocitv  of  12  feet 


1 42  DYNAMICS. 

per  minute,  how  many  grains  of  light  from  the  sun  would  be 
sufficient  to  stop  it  ? 

Ex.  5.  A  train  of  cars  moving  with  a  velocity  of  20  miles 
an  hour,  had  been  gone  three  hours,  when  a  locomotive  was 
dispatched  in  pursuit,  with  a  velocity  of  25  miles  an  hour ;  in 
what  time  did  the  latter  overtake  the  former? 

Ex.  6.  Had  the  trains  in  Ex.  5  started  together  and  moved 
in  opposite  directions  around  the  earth,  21,840  miles,  in  whai 
time  would  they  meet? 

Ex.  7.  Suppose  it  to  be  12  minutes  past  noon  by  a  clock, 
in  how  long  a  time  will  the  hour  and  minute  hands  of  the 
clock  be  together  ? 

Ex.  8.  The  daily  motion  of  Mercury  in  his  orbit  is  4°.09239  ; 
that  of  Venus  1°.60216;  that  of  the  earth  0°.98563  :  what  are 
the  intervals  between  the  epochs  at  which  Mercury  and  Venus 
respectively  will  be  in  the  same  direction  from  the  sun  as  the 
earth  ? 

Ex.  9.  A  man  being  caught  in  a  shower  in  which  the  rain 
fell  vertically,  ran  with  a  velocity  of  12  feet  per  second.  He 
found  that  the  drops  struck  him  in  the  face,  and  estimated  that 
the  apparent  direction  of  the  drops  made  an  angle  of  10°  with 
the  vertical  line.     What  was  the  velocity  of  the  drops  ? 

Ans.  68  feet. 

Ex.  10.  When  the  path  of  the  earth  in  its  orbit  is  per  pen 
dicular  to  a  line  drawn  from  a  star  to  the  earth,  the  path  of  the 
light  from  the  star  appears  to  make  an  angle  of  20",445  with 
the  perpendicular  to  the  path  of  the  earth.  The  velocity  of 
the  earth  being  68,180  miles  per  hour,  what  is  the  velocity 
of  light? 


CHAPTER  II. 

IMPACT     OF     BODIES. 

240.  Def.  When  two  bodies  in  motion  impinge,  if  their  cen- 
ters of  inertia  move  in  the  same  straight  line  perpendicular  to 
a  plane  tangent  to  the  bodies  at  their  point  of  contact,  the  im- 
pact is  said  to  be  direct  and  central. 

If  the  straight  line  described  by  the  center  of  inertia  of  one 
of  the  bodies  is  not  perpendicular  to  the  tangent  plane,  the  im- 
pact is  said  to  be  oblique. 

In  the  cases  discussed  the  bodies  will  be  supposed,  spherical 
and  of  uniform  density. 

241.  Def.  When  the  bodies  impinge,  they  exert  a  mutual 
but  varying  pressure  during  the  interval  between  contact  and 
separation,  an  interval  of  time  which  is  generally  very  short, 
and  we  suppose  them  to  suffer  a  degree  of  compression,  by 
which,  during  a  portion  of  this  interval,  their  centers  will  ap- 
proach each  other,  and  during  the  remaining  portion  will  re- 
cede by  the  action  of  an  internal  force  tending  to  restore  them 
to  their  original  form.  The  force  urging  the  approach  of  their 
centers  is  called  the  force  of  comj)ression ;  the  opposing  force 
causing  them  to  separate  again  is  called  \\\q  force  of  restitution 
or  elasticity.  The  ratio  of  the  force  of  restitution  to  that  of 
compression  is  called  the  modulus  of  elasticity. 

When  this  ratio  is  unity,  or  the  force  of  restitution  is  equal 
to  that  of  compression,  the  bodies  are  perfectly  elastic ;  when 
it  is  zero,  or  the  force  of  restitution  is  nothing,  they  are  inelastic. 
If  the  value  of  the  ratio  is  intermediate  between  zero  and  unity 
the  bodies  are  imperfectly  elastic. 

242.  Def.  If  the  bodies  suffer  no  compression,  they  are  call- 
ed hard ;  if,  when  compressed,  they  exert  no  force  to  recover 
their  original  form,  they  are  called  soft. 

There  are  no  known  bodies  either  perfectly  elastic  or  per- 


144  DYNAMICS. 

fectly  inelastic,  but  these  states  may  be  considered  as  limits  to 
the  various  degrees  of  elasticity  presented  in  nature. 

243.  Prop.  To  determine  the  velocity  of  two  inelastic  bodies 
after  direct  impact. 

Since  the  bodies  are  inelastic,  the  force  of  restitution  is  zero, 
and  the  bodies  will  move  on  together  with  a  common  velocity. 

1°.  Let  ml  and  m2  be  the  masses  or  bodies  moving  in  the 

©„  same  direction  with  the  velocities 
mz 
>■ Q >-    vl  and  v2  {v^v^,  and  v  their 
common  velocity  after  impact. 
If  F  and  F'  are  the  forces  which  impress  on  the  bodies  their 
respective  velocities,  then  (38) 

F^mjVj,  and  F'=m2v2, 
and  their  resultant,     F+F'=mlvl  +m2v2. 

After  impact  the  bodies  move  on  together  as  one  mass,  and 
its  momentum  (ml-\-m2)v  must  be  a  measure  of  the  force 
F+F'. 

.-.  {ml+m2)v=mxv1-\-m2v2, 

mlv1+m2v2 

or  v= ; .  (46) 

ml+m2 

2°.  If  the  bodies  move  in  opposite  directions,  the  resultant 
of  the  forces  will  be  F— F'. 

.'.  F— F'=m1v1  —  m2v2  =  (m1+m2)v, 

mlvl—m2v2 
or  u=  , — —.  (47) 

The  same  result  will  be  obtained  by  changing  the  sign  of 
v2  in  (46). 

Hence  the  velocity  of  two  inelastic  bodies  after  impact  is 
equal  to  the  algebraic  sum  of  their  momenta  before  impact,  di- 
vided by  the  sum  of  their  masses. 

Cor.  1.  If  two  bodies  move  in  opposite  directions,  with  veloci- 
ties reciprocally  proportional  to  their  masses,  they  will  rest  after 
impact. 

For  ml  :  m2—va  :  v, 

gives  m  v1=m2v2, 

which,  substituted  in  (47),  gives  v=0. 


IMPACT     OF     BODIES.  145 

Con.  2.  If  m2  is  at  rest  before  irr pact,  v2=0,  and 

m ,  v , 


?n1  +m2  ' 
and  if,  at  the  same  time,  the  masses  are  equal, 

v=r 

Cor.  3.  If  the  masses  are  equal,  and  move  in  the  same  di 
reotion, 

f  m  opposite  directions, 


2 

244.  Prop.  In  the  impact  of  inelastic  bodies  there  is  a  loss  of 
living  force,  and  this  loss  is  equal  to  the  sum  of  the  living  forces 
due  to  the  velocities  lost  and  gained  by  the  bodies  respectively. 

For  the  living  force  before  impact  ^m^l+m^l  (Art.  224), 
and        "       "  "     after        "       =  (ml+m2)v\ 

.'.  the  loss  of  living  force  by  impact  is 
m1v2+m.2v22—(m1+m2)v*=m1v2-{-?n;iv22  —  2(m1+m2)v1+ (m^+m^v* 
(46),  =mlv2 +m2v22  —  2(mlvl+m2v2)v-\-(m1+m2)v'1 

=ml(vi  —  vy+m2(v2—v)\ 
which  is  necessarily  positive,  and  in  which  vx—v  is  the  veloci- 
ty lost  by  m,,  and  v—va  the  velocity  gained  by  m2. 

From  this  proposition  it  appears  that  in  machinery  made  of 
inelastic  materials  all  abrupt  changes  of  motion  are  attended 
with  a  loss  of  living  force,  by  which  loss  the  efficiency  of  the 
machinery  is  impaired. 

245.  Prop.  To  find  the  velocities  of  two  imperfectly  elastic 
bodies  after  direct  impact. 

Let  ?n1  and  m2  be  the  masses,  vx  and  v2  their  velocities  be- 
fore impact,  and  v'lt  v'2  their  velocities  after  impact. 

The  bodies  being  elastic,  will  suffer  compression.  Let  v  be 
their  common  velocity  at  the  instant  of  greatest  compression 
or  when  the  distance  between  their  centers  is  least.  Then  the 
velocity  lost  by  ml  at  this  instant  wil  be  vl—  v. 

K 


14G 


DYNAMICS. 


Let  e  be  the  modulus  of  elasticity,  or  the  ratio  of  the  torce 
of  restitution  to  that  of  compression.  Since  these  forces  are 
pioportional  to  the  velocities  they  generate  or  destroy  in  the 
same  mass,  the  velocity  destroyed  in  m  l  by  the  force  of  resti- 
tution will  be  e(v1—v). 

Hence  the  whole  velocity  lost  by  m,  will  be 
vl—  v+e(v1—  v)  =  (l+e)  (v^—v). 
This,  subtracted  from  the  velocity  of  ml  before  impact,  will 
give  its  velocity  after  impact,  or 

v\  =  v1  —  (1+e)  (p  1—v)=v—e(v  1—v).  (a) 

In  like  manner,  the  velocity  gained  by  m2  during  compres- 
sion will  be  v— v2,  and  the  velocity  gained  by  the  force  of 
restitution  e(v—v2). 

Hence  the  whole  velocity  gained  by  mQ  will  be  (1+e) 
\V— v2).  This,  added  to  the  velocity  before  impact,  will  give 
the  velocity  after  impact,  or 

v'2  =  v2  +  (l+e)  (v—v2)—v+e(v—v2).  (b) 

Substituting  in  (a)  and  (b)  the  value  of  v  (46),  and  reducing, 


m1+m2  7n1+?n2 


(48) 


2        ml+?n2  ?n1+m2 

As  in  Art.  243,  if  the  bodies  move  in  opposite  directions,  we 
must  change  the  sign  of  v2,  or,  if  one  of  them  be  at  rest  before 
impact,  make  u^O.  Also,  if  we  put  e=l,  the  results  will  be 
those  for  perfectly  elastic  bodies,  or  make  e=0,  the  result  will 
be  that  for  inelastic  bodies. 

Cor.  1.  If  the  bodies  are  perfectly  elastic,  their  relative 
velocities  before  and  after  impact  are  the  same.  For,  making 
£=1  in  (a)  and  (b),  and  subtracting  the  latter  from  the  former, 
we  have 

v'l  —  v'2=v2—vl.  (50) 

Cor.  2.  In  the  impact  of  bodies  no  motion  is  lost. 

For,  multiplying  (48)  by  ml,  and  (49)  by  m2,  adding  and 
reducing,  we  have 

mlv'l+m2v'2—mlvl-\-m2v2,  (51) 


IMPACT     OP     BODIES.  1 47 

in  which  the  first  member  is  the  sum  of  their  momenta  iftei 
impact,  and  the  second  member  the  sum  of  their  momenta  be- 
fore impact. 

Cor.  3.  If  the  bodies  are  perfectly  elastic  and  equal,  they 
will  interchange  velocities  by  impact. 

For,  making  ?n1=?n2  and  £=1  in  (48)  and  (49),  we  have 

v2=i(vi+v2)  +  2(vi—v2)  =  vi- 
Cor.  4.  The  velocity  which  one  body  communicates  to  an 
ithev  at  rest,  when  perfectly  elastic,  is  equal  to  twice  the  ve- 
locity of  the  former  divided  by  one  plus  the  ratio  of  the  masses. 
Making  in  (49)  m2=rm1,  v2—Q,  and  e=l,  we  obtain 

2     r+1" 

246.  Prop.  When,  in  a  series  of  n  perfectly  elastic  bodies 
whose  masses  are  in  geometrical  progression,  the  first  impinges 
directly  against  the  second  at  rest,  the  second  against  the  third, 
and  so  on,  to  find  the  velocity  of  the  nth  body. 

Let  m,  rm,  r^m  ....  rn~1m  be  the  bodies,  and  v  the  velocity 

of  m.     By  Cor.  4,  Art.  245, 

2v 
the  velocity  of  the  second  will  be 


third 
"       fourth 

"       nth 


r+1' 

2v       1  2*v 


1+r'l+r     (1+r)2' 
22t>        1  2sv 


'(l+r)s,l+r     (1+r)3' 

2,-1u 


(1+r)'-1* 
Cor.  1.  The  momentum  of  the  nth  body  is 
T'lv  (   2r 

Cor.  2.  If  the  bodies  are  equal,  r=l,  and  the  velocity  of  the 
last  equals  v,  the  velocity  of  the  first.  If  all  are  in  contact  ex 
cept  the  first  before  impact,  all  except  the  last  will  remain  in 
contact  after  impact,  and  the  last  will  move  off  with  the  ve- 
locity of  the  first. 


1 48  DYNAMICS. 

247.  Prop.  The  motion  of  the  common  center  of  gravity  vf 
two  bodies  after  direct  impact  is  the  same  as  before  impact. 

Let  v  be  the  velocity  of  the  common  center  of  gravity  of  the 
bodies  m1  and  m2,  moving  in  the  same  direction  with  veloci- 
ties vl  and  v2  respectively  before  impact,  and  v'  the  velocity 
of  their  common  center  of  gravity  after  impact.  Let  xx,  xt> 
x  be  the  distances  respectively  of  the  centers  of  gravity  of  m, 
and  m2,  and  of  their  common  center  of  gravity  from  any  fixed 
point  in  their  line  of  motion  at  any  instant ;  x'l,  x'2,  xl  the  same 
quantities  after  an  interval  t,  so  that  (36) 

x'2=x2+v2t, 
x'  =x  -\-vt. 
By  (29),  Art.  105, 

(m1+)n2)x  =m1xl+m2x2,  (a) 

(ml+m2)x'=m1x'1-\-m2x'2.  (b) 

Substituting  in  (b)  the  values  of  x',  x'2,  x\  above,  and  subtract 
ing  (a)  from  the  result,  we  have 

m1+m2  v 

which  is  the  same  as  the  velocity  of  two  inelastic  bodies  after 
impact,  and  therefore  equal  to  the  velocity  of  their  common 
center  of  gravity  after  impact,  since  the  masses  move  on  to- 
gether. 

If  the  bodies  are  elastic,  and  v\ ,  v'2  are  their  velocities  after 
impact,  then 

x'1=x1+v'lt, 

X  2         X  2~T~V  2t, 

x'  =x  -\-v't; 
from  which  we  deduce,  as  before, 

m1+m2  v  ' 

for  the  velocity  of  the  common  center  of  gravity  of  elastic 
bodies  after  impact.  Now  the  denominators  of  (c)  and  (d)  are 
the  same,  and  by  (51)  the  numerators  are  equal ; 


IMPACT     OF     BODIES. 


149 


or  ;he  velocity  of  the  common  center  of  gravity  is  unchanged 
by  impact. 

248.  Schol.  This  proposition  is  only  a  particular  case  of  a 
general  principle  in  Mechanics,  denominated  the  conservation 
of  the  motion  of  the  center  of  gravity.  The  principle  consists 
in  this,  that  the  mutual  action  of  the  several  bodies  or  parts  of 
a  system  upon  each  other  can  produce  no  change  in  the  mo- 
tion of  the  center  of  gravity  of  the  entire  system. 

249.  Def.  If  a  body  impinge  on  a  surface,  the  angle  which 
its  path,  before  impact,  makes  with  the  perpendicular  to  the 
surface  at  the  point  of  impact  is  called  the  angle  of  incidence, 
and  the  angle  which  its  path,  after  impact,  makes  with  the 
same  perpendicular  is  called  the  angle  of  reflection. 

250.  Prop.  To  determine  the  motion  of  a  smooth  inelastic 
body  after  oblique  impact  upon  a  smooth,  hard,  and  fixed  plane. 

Let  the  body  m  impinge  on  the  plane 
AC  at  B,  with  the  velocity  v,  making 
the  angle  of  incidence  mBN—6. 

The  component  of  the  velocity  v 
parallel  to  the  plane  (39)  is  v.  sin.  6 ; 
and  this  velocity  will  not  be  changed 
by  impact,  since  the  body  and  plane  are  smooth.  The  com- 
ponent of  v  perpendicular  to  the  plane,  viz.,  v.  cos.  0,  will  be 
destroyed  by  the  plane,  and,  since  the  body  and  plane  are  in- 
elastic, there  will  be  no  vertical  velocity  after  impact.  Hence 
the  body  will  slide  along  the  plane  with  the  velocity  v.  sin.  0. 

251.  Prop.  To  determine  the  motion  of  an  elastic  body  after 
oblique  impact  upon  a  smooth,  hard,  and  fixed  plane. 

Let  the  body  m  impinge  on  the  plane 
RS  at  B,  with  the  velocity  v,  making  the 
angle  of  incidence  PBN=0.  Let  PB  rep- 
resent the  velocity  v  before  impact.  Draw 
PR  and  PN  perpendicular  and  parallel  to 
the  plane  RS.  PN=RB=u.  sin.0  is  the 
component  of  the  velocity  parallel  to  the  plane,  and  is  not  af- 
fected by  the  impact.  PR  =  NB=u.  cos.  0,  the  component  ot 
the  velocity  perpendicular  to  the  plane,  will  be  destroyed  by 


150 


DYNAMIC 


the  plane.  But  the  body  being  elastic,  the  force  of  restitution 
will  give  it  a  velocity  e.v.  cos.  6  in  the  direction  BN.  Take 
BM=e.t\  cos.  6,  BS=u.  sin.  0,  and,  completing  the  parallelo- 
gram, draw  BQ.  BQ/rc'  is  the  direction,  and  BQ,  the  velocity 
of  the  body  after  impact.     Now 

BQ2=v'2=:BM2+MQ2 

=e.V  cos.a  0+wa  sin.9  0 


also, 


BQ=v'=v  Ve*  cos.s  0+sin.a  0 ; 
MQ_  PN  _tan.0 
tan-     _MB_lNB-     e    ' 


Cor.  If  £=1,  6=6',  and  v  =  v',  or  if  the  body  be  perfectly 
elastic,  the  angle  of  incidence  equals  the  angle  of  reflection, 
and  the  velocity  is  the  same  after  as  before  impact.  If  e=0, 
tan.0'=oD,  and  6' =90°. 

252.  Prop.  To  determine  the  direction  in  which  a  body  of 
given  elasticity  must  be  projected,  in  order  that  after  reflection 
from  a  given  plane  it  may  pass  through  a  given  point. 

Let  MB  be  the  given  plane,  A  the 
position  of  the  body  before  projection, 
D  the  point  through  whi'ch  it  is  re- 
quired to  pass  after  reflection,  and  e 
the  modulus  of  elasticity  of  the  body, 
supposed  known. 

Draw  AMC  perpendicular  to  the 
plane,  and  take  AM  :  MC  =  1  :  e 
Draw  C  D  cutting  the  plane  in  B,  and  join  A  B.  AB  is  the 
direction  in  which  the  body  must  be  projected,  and  AB,  BD 
will  be  its  path. 

Since  e.AM=MC, 

MB      MB 

eTAM-MC 

tan.  MAB 


or 


or 


£ 

tan.0 


=tan.  MCB ; 


=tan.0\ 


as  in  Art.  251. 


IMPACT     OF     BODIES. 


151 


253.  Prop.   The  modulus  of  elasticity  is  equal  to  the  ratio  oj 
the  relative  velocities  of  the  bodies  after  and  before  impact. 
For,  eliminating  v  from  (a)  and  {b)  {Art.  245),  we  have 

(52) 


u,  —  v2 

254.  Schol.  Bodies  suspended  by  fine  cords,  and  made  tc 
oscillate  like  a  pendulum,  acquire  velocities  at  the  lowest  point 
proportional  to  the  chords  of  the  arcs  of 
descent,  and  will  rise  through  arcs  whose 
chords  are  proportional  to  the  velocities 
impressed  upon  them  at  the  lowest  point 
of  the  arcs.  Also,  if  the  arcs  be  small, 
the  times  of  descent  will  be  equal,  so 
that  bodies  descending  through  arcs  of 
different  lengths  will  impinge  at  the 
lowest  points.  If,  therefore,  two  spher- 
ical bodies  of  the  same  material  be  made  to  descend,  in  the 
manner  described,  through  arcs  of  given  length,  and  the  arcs 
through  which  they  rise  after  impact  be  measured,  the  veloci- 
ties vx,v2,v\,  v'3  will  be  known,  and  these,  substituted  in  (52), 
will  give  e  the  modulus  of  elasticity. 

The  following  table  exhibits  the  results  of  experiments,  per- 
fect elasticity  being  unity : 


2  1  0  1   2 


Substances. 


Moduli. 


Substances. 


Moduli. 


Glass.  .  .  . 
Hard-baked  clay 
Ivory.  .  .  . 
Limestone  .  . 
Steel,  hardened 
Cast  iron 
Steel,  soft    .     . 


0.94 
.89 
.81 
.79 
.79 
.73 
.67 


Bell-metal 

Cork 

Bras9 

Lead 

Clay,  just  malleable  by  the 
hand 


0.67 
.05 
.41 
.20 

.17 


255.    EXAMPLES. 

Ex.  1.  Two  inelastic  bodies,  weighing  12  lbs.  and  7  lbs.  re- 
spectively, move  in  the  same  direction  with  velocities  of  8  feet 
and  5  feet  in  a  second.  Find  the  common  velocity  after  impact ; 
also  the  velocity  lost  by  one,  and  tVat  gained  by  the  other. 

Ex.  2.  A  mass  w,,  with  a  velocity  11,  impinges  on  m2  moving 
in  the  opposite  direction  with  a  velocity  of  5;  by  impact  ml 


1 52  DYNAMICS. 

loses  one  third  of  its  momentum.  What  are  the  relative  mag- 
nitudes of  ml  and  m21 

Ex,  3.  m,,  weighing  8  lbs.,  impinges  on  m2,  weighing  5  lbs., 
and  moving  in  m^s  direction  with  a  velocity  of  9;  by  impact 
m2s  velocity  is  trebled.  What  was  m,'s  velocity  before  im- 
pact? 

Ex.  4.  Two  bodies  mx  and  m2  are  moving  in  the  same 
straight  line  with  velocities  u,  and  v2.  Find  the  velocity  of 
each  after  impact  when  6??i1=5m2,  vl=l,  4o1+5t>2=0,  and 

C  3* 

Ex.  5.  Two  bodies  m,  and  m2  are  perfectly  elastic  and 
move  in  opposite  directions;  in1  is  treble  of  ma,  but  m2's 
velocity  is  double  that  of  m1.  Determine  their  motions  after 
impact. 

Ex.  6.  There  is  a  row  of  perfectly  elastic  bodies  in  geomet- 
rical progression  whose  common  ratio  is  3 ;  the  first  impinges 
on  the  second,  which  transmits  its  velocity  to  the  third,  and  so 
on ;  the  last  body  moves  off  with  -g\  the  velocity  of  the  first. 
What  was  the  number  of  bodies  ? 

Ex.  7.  m1(=3m2)  impinges  on  m2  at  rest;  yn/s  velocity 
after  impact  is  f  of  its  velocity  before  impact.  Required  the 
value  of  e,  the  modulus  of  elasticity. 

Ex.  8.  Two  bodies  mi  and  m„,  whose  elasticity  is  f,  mo-ing 
in  opposite  directions  with  velocities  25  and  16  respectively, 
impinge  directly  upon  each  other.  Find  the  distance  between 
them  4t\  seconds  after  impact. 

Ex.  9.  At  what  angle  must  a  body  whose  elasticity  is  \  be 
incident  on  a  perfectly  hard  plane,  that  the  angle  made  by  its 
path  before  and  after  impact  may  be  a  right  angle? 

Ex.  10.  A  ball  whose  elasticity  is  e,  projected  from  a  given 
point  in  the  circumference  of  a  circle,  after  two  reflections 
from  the  interior  of  the  circle,  returns  to  the  same  point.  Re- 
quired the  angle  8  made  by  the  direction  of  projection  with  tr>e 
radius  at  the  given  point. 

3 

e2 
Ans.  Tan.  6=    .-—  -3. 


CHAPTER   III. 

MOTION    FROM    THE    ACTION    OF    A    CONSTANT    FORCE. 

256.  By  the  definition  of  a  constant  force  (Art.  222),  the 
velocities  generated  in  equal  successive  intervals  of  time  by 
the  action  of  the  force  are  all  equal,  and  the  increment  of  ve- 
locity in  a  unit  of  time  is  a  measure  of  the  force.  Hence  the 
velocity  is  the  same  at  no  two  successive  instants,  and,  if  the 
body  or  point  move  from  rest,  will  increase  uniformly,  or  be 
uniformly  accelerated. 

By  the  velocity  acquired  in  t  seconds  is  meant  the  space 
over  which  the  body  would  pass  in  the  second  next  succeed- 
ing the  t  seconds  if  the  velocity  should  remain  the  same  during 
this  second  as  it  was  at  the  end  of  the  t  seconds  ;  or  the  space 
described  by  the  body  in  the  interval  between  t  seconds  and 
t+1  seconds,  if,  at  the  end  of  t  seconds,  the  force  should  cease 
to  act.  Putting/  for  the  force,  and  v  for  the  velocity  acquired 
in  the  time  t,  we  have,  as  before, 

v=ft.  (53) 

257.  PRor.  To  find  the  space  in  terms  of  the  force  and  time 
when  a  body  moves  from  rest  by  the  action  of  a  constant  force. 

Let  f  be  the  force,  and  s  the  whole  space  described  in  the 

time  t,  and  let  t  be  divided  into  n  equal  parts,  each  =-.     The 

intervals  reckoning  from  the  commencement  of  the  time,  will 
be 

t   2t  3t  4d  (n—\)t  nt 

~i        >        j        t  CZtC.f ,        . 

n   n    n    n  n         n 

By  (53),  the  velocities  at  the  end  of  these  intervals  will  be> 

1  n,J  n'J  n'J  n  '  '  *  '  J'      n     ,J'n' 


154  DYNAMICS. 

If  now  the  body  moved  uniformly  during  each  interval  - 

with  the  velocity  it  had  at  the  beginning  or  end  of  this  interval, 
the  spaces  described  during  the  intervals  respectively,  would 
be  equal  to  the  product  of  this  uniform  velocity  by  the  interval 
(37),  and  the  whole  space  described  would  be  equal  to  the  sum 
of  these  partial  spaces. 

If,  therefore,  the  body  moved  uniformly  during  each  interval 

-  with  the  velocity  it  had  at  the  beginning  of  the  interval,  we 

should  have 

„  f      r2f      „3f       „  „(n-l)*8' 

s=0+f.~+f—+f~+,  &c,  .  .  .  f> ~, 

J  n     J  n      J  n  J        n 

f 
=/.-{l+2  +  3+,  &c,  .  .  .  (w-1)}, 


/ 


n 

t~  n(?i—l) 

n*'      2      ' 


f  f 

=f-2-f'¥n  ^ 

But  if  the  body  moved  uniformly  during  each  interval  - 

with  the  velocity  it  had  at  the  end  of  the  interval,  we  should 
have 

,  *a  ,    ,  M         3t\  (n-l)f         nt 

J  n     -    n      J    n  J        n  J    n 

f 
=/.^(l+2+3+,&c, n), 

f  n(n+l) 
~~J'nv      2      ' 

f         f 

Since  the  velocity  is  uniform  during  no  sensible  interval,  the 
true  value  of  s  will  lie  between  the  two  quantities  (a)  and  (b), 
however  small  each  interval  may  be,  or  however  large  n  may 
be.  But  when  n  becomes  indefinitely  large,  the  last  terms  in 
(a)  and  (b)  vanish,  and  (a)  =  (b). 


ACTION  OF  A  CONSTANT  FORCE.        155 

.-.  s=±f?  and  sx t\  (54) 

Hence  the  space  described  from  rest  by  a  oody  from  the 
action  of  a  constant  force  is  equal  to  half  the  product  of  the  force 
by  the  square  of  the  number  of  seconds,  and  the  spaces  vary  as 
the  squares  of  the  times. 

258.  Prop.  To  determine  the  space  in  terms  of  the  force  and 
velocity ;  also  in  terms  of  the  time  and  velocity. 

1°.  Eliminating  t  from  (53)  and  (54),  we  have 

v2 
s=-7,  orsx^.  (55) 

3 

2°.  Eliminating/  from  (53)  and  (54),  we  obtain 

s=^vt,  or  socvt.  (56) 

Cor.  The  space  described  in  any  time  by  a  body  moving 
trom  rest  by  the  action  of  a  constant  force,  is  half  that  it  would 
describe  in  the  same  time  if  it  moved  uniformly  with  the  ac- 
quired velocity. 

For  the  space  s„  described  in  the  time  t,  with  a  uniform  ve- 
locity v,  is,  by  (37), 

s1=vt, 
which,  compared  with  (56),  gives 

259.  Prop.  To  find  the  space  described  by  a  body  in  the  last 
n  seconds  of  its  motion. 

The  space  described  in  the  time  t  (54)  is 

sx=\ft\  (a) 

The  space  described  in  the  time  (t—n)  is 

Subtracting  (b)  from  (a),  we  have,  for  the  space  s  described 
in  the  last  n  seconds, 

g=sx-sa=%f(2nt-n3).  (57) 

Cor.  If  the  space  described  in  the  m  seconds  next  preceding 
the  last  n  seconds  is  required,  we  have,  for  the  space  s3,  de- 
scribed in  the  time  t—  (n+m)  seconds,  s a=^f(t—n—my,  which, 
subtracted  from  (b),  gives  for  the  space  s  required, 

s=s2-s.J=$f(2mt-2mn-mu).  (58) 


156  DYNAMICS. 

2G0.  Prop.  A  body  being  projected  with  a  given  velocity  in 
the  direction  in  which  a  constant  force  acts,  to  find  the  velocity 
of  the  body  at  the  end  of  a  given  time,  and  the  space  described  in 
that  time. 

By  Art.  234,  the  velocity  due  to  the  joint  action  of  the  im- 
pulsive and  constant  forces  will  be  equal  to  the  sum  or  differ- 
ence of  the  velocities  due  to  each,  according  as  they  act  in  the 
same  or  opposite  directions.  If,  therefore,  vJ  be  the  velocity 
of  projection,  the  whole  velocity  v,  at  the  end  of  the  time  t, 
will  be 

»=Pt±/t  (59) 

In  the  same  manner,  the  space  due  to  the  joint  action  of  the 
forces  will  be  equal  to  the  sum  or  difference  of  the  spaces  due 
to  each,  or 

s=v1t±^ff.  (60) 

261.  Prop.  A  body  being  projected  with  a  given  velocity  in 
the  direction  in  which  a  constant  force  acts,  to  find  its  velocity 
when  it  has  passed  through  a  given  space.  • 

Let  s  be  the  given  space,  and  h  the  space  through  which  the 
body  must  pass  to  acquire  the  velocity  vx  by  the  action  of  the 
constant  force.     Then  (55)  . 

v\  =  2fh, 
and  for  the  space  (h±s) 

v*=2f(h±s), 
=2fh±2fs, 
=v\±2fs,  (61) 

the  signs  to  be  taken  as  in  the  last  proposition. 

262.  Prop.  When  a  body  is  projected  in  a  direction  opposilt 
to  that  in  which  a  constant  force  acts,  the  velocity  acquired  in 
returning  to  the  point  of  departure  is  equal  to  the  velocity  of 
projection. 

If  t),  be  the  velocity  of  projection  in  the  direction  AB,  from 
(61)  we  have 

*=~w  (a) 


ACTION     OF     A     CONSTANT     FORCE.  157 

Now  the  actual  velocity  v  of  the  body  is  continual- 
ly diminished  by  the  action  of  the  force/,  and  when 
the  body  has  reached  its  greatest  distance  from  A,  as 
B,  v=0.     Hence  (a)  gives 

s=^jr,  or  v1=V2fs. 

But  the  velocity  v  acquired  in  moving  from  rest 
through  BA=s  is,  by  (55), 

v=  Vlfs.  , 

Cor.  1.  The  velocity  of  the  body  at  any  given  dis- 
tance from  A  is  the  same  in  going  and  returning. 

For  in  the  expression  (a),  since  u,  and  2f  are  constant,  v  is 
always  the  same  for  the  same  value  of  s=AC. 

If  u>Uj,  s  is  negative,  or  on  the  opposite  side  of  A  from  B. 

2v 
Cor.  2.  The  whole  time  of  flight  T=-^.     Making  v=0  in 

v 
(59),  we  have  t=-j-,  when  the  body  is  at  B. 


But,  in  returning,  it  acquires  the  velocity  vl=ft. 

Hi 

/' 


V 

Hence  the  time  of  return  is  t—-^r, 


and  the  whole  time  2£=T= 

263.  Schol.  1.  There  is  no  known  instance  in  nature  of  a  force 
which  is  constant.  The  law  of  Universal  Gravitation  is,  that 
every  particle  of  matter  attracts  every  other  particle  with  a 
force  which  varies  directly  as  the  mass  of  the  attracting  particle, 
and  inversely  as  the  square  of  the  distance.  A  sphere  of  uni- 
form density,  or  one  whose  density  is  the  same  at  equal  distan- 
ces from  the  center,  attracts  a  body  exterior  to  it  as  if  the  mat- 
ter of  the  sphere  were  collected  at  its  center,  and  with  a  force 
varying  inversely  as  the  square  of  the  distance  of  the  body  from 
the  center,  but  a  body  or  particle  in  the  interior  with  a  force 
varying  directly  as  its  distance  from  the  center. 

Regarding  the  earth  as  a  sphere,  this  is  the  law  of  the  earth's 


158  DYNAMICS. 

attraction  for  bodies  exterior  to  it ;  but  for  all  small  distances 
above  the  surface  the  intensity  of  gravity  may  be  considered 
constant,  since  at  a  distance  of  one  mile  above  the  surface  the 
actual  diminution  of  gravity  is  only  li)77'i:!!)1,  or  about  the  2000th 
part  of  that  at  the  surface  ;  a  variation  too  small  to  affect  sens- 
ibly the  circumstances  of  the  motion  of  a  falling  body  com 
puted  on  the  hvpothesis  that  the  force  suffers  no  variation  at 
all. 

In  reality,  the  force  by  which  a  body  is  drawn  toward  the 
earth  is  equal  to  the  sum  of  the  attractions  of  each  for  the 
other ;  but  when  the  mass  of  the  body  is  inconsiderable  with 
regard  to  the  mass  of  the  earth,  the  effect  of  the  former  is  in- 
sensible, and  the  accelerations  of  all  bodies  of  moderate  size 
are  the  same.  Within  these  limits,  then,  gravity  may  be  taken 
as  a  constant  or  uniformly  accelerating  force,  whatever  be  the 
mass. 

264.  Schol.  2.  Representing  the  intensity  of  gravity  at  the 
surface  of  the  earth,  as  before,  by  g,  we  have,  from  (54), 
s=\gf,  and  if  t=l,  s=hg,  or  g=2s  from  which  it  appears 
that  the  acceleration  is  equal  to  twice  the  space  described  ir. 
the  unit  of  time.  It  has  been  found  by  experiment  that  the 
space  through  which  a  body  falls  freely  in  one  second  in  the 
latitude  of  New  York  is  equal  16.0799  feet,  or  16^  feet  nearly. 
Hence  g— 32. 1598  feet,  or  32£  feet  nearly.  Hence,  also,  all 
the  relations  between  the  space,  time,  and  velocity  due  to  the 
action  of  a  constant  force  are  true  in  relation  to  the  action  of 
gravity  near  the  earth's  surface.  Collecting  these,  and  substi- 
tuting g  for  /,  we  have  the  following  relations,  supposing  no 
resistance  from  the  air : 

^=W=~=UV,  (54') 

•  2s         

v=  gt=~r  =  V2gs,  (53') 


v       2s         /2s  ,fflrt 

t=  -   =—  ==V— ,  (56') 

S        v        V    g 

0*5') 


g         V  V     g 

v    _2s  _o5 
1   =F=2? 


ACTION     OF     A     CONSTANT     FORCE.  159 

Also,  when  a  body  is  projected  with  a  velocity  v1  vertically 
upward  or  downward,  (260)  and  (261), 

s=v1t±lgt\  (60') 

tf=v\  ±2gs.  (61') 

265.    EXAMPLES. 

Ex.  1.  A  body  has  been  falling  11  seconds.  Find  the  space 
described  and  the  veloctiy  acquired. 

By  (54'),         s=i£i'=16TVXl21  =  1946TV  feet. 
By  (53'),         v=  gt=32}  X    11=  353|  feet. 

Ex.  2.  Find  the  time  in  which  a  falling  body  would  acquire 
a  velocity  of  500  feet,  and  the  height  from  which  it  must  fall. 

^  v       500  . 

By  (56'),    t=  -  =—-  =  15.544  seconds. 

„.     ,      ,  v2      250000 

By  (54'),   s=—  =-^-=3886  feet  nearly, 

or  *=4*u=£x  15.544X500=3886  feet,  as  before. 

Ex.  3.  What  is  the  velocity  acquired  by  a  body  in  falling 
450  feet  ?  and  if  the  body  weigh  10  tons,  what  is  the  momentum 
acquired  ? 

Ans.  v  =170.14  feet. 

M=3811136lbs. 

Ex.  4.  A  body  had  fallen  through  a  height  equal  to  one 
quarter  of  a  mile.  What  was  the  space  described  by  it  in  the 
last  second? 

Ans.  s=275  feet. 

Ex.  5.  A  body  had  been  falling  15  seconds.  Compare  the 
spaces  described  in  the  seventh  and  last  seconds. 

Ex.  6.  A  body  had  been  falling  12.5  seconds.  What  was 
the  space  described  in  the  last  second  but  5  of  its  fall? 

Ex.  7.  The  space  described  by  a  body  in  the  fifth  second 
of  its  fall  was  to  the  space  described  in  the  last  second  but  4, 
as  1  to  6.     What  was  the  whole  space  described? 

Ans.  s=15958.69  feet. 

Ex.  8.  A  body  is  projected  vertically  downward  with  a  ve- 


1 60  DYNAMICS 

locity  of  100  feel.     What  is  its  velocity  at  the  end  of  5  sec« 
onds?  ■   =    -  .'  o. 

Ex.  9.  A  body  is  projected  vertically  upward  with  a  veloci- 
ty of  100  feet.  Find  its  velocity  at  the  end  of  5  seconds,  and 
its  position  at  the  end  of  8  seconds. 

Ans.  v=— 60f  feet. 
s=—  229^  feet. 

Ex.  10.  A  body  is  dropped  into  a  well  and  is  heard  to  strike 
the  bottom  in  4  seconds.  What  is  the  depth  of  the  well,  the 
velocity  of  sound  being  1130  feet? 

Ans.  231  feet. 

Ex.  11.  A  body  is  thrown  vertically  upward  with  a  veloci- 
ty »,.  Find  the  time  at  which  it  is  at  a  given  height  h  in  its 
ascent. 

If  t  be  the  time  required,  (60)  gives 
h=v1t— \gf\ 

,     2v.     2h 

whence  t—t — H =0, 

g       g 


v,±Vv*—2gh 

or  t  =— — . 

g 

The  lower  sign  gives  the  time  when  the  body  is  at  the 

height  h  in  its  ascent,  and  the  upper  in  its  descent. 

Ex.  12.  A  body  is  projected  vertically  upward,  and  the  in- 
terval between  the  times  of  its  passing  a  point  whose  height  is 
h  in  its  ascent  and  descent  is  21.  Find  the  velocity  v  of  pro- 
jection, and  the  whole  time  T  of  its  motion. 

Ex.  13.  A  body  whose  elasticity  is  £  is  projected  vertically 
upward  to  the  height  h  above  a  hard  plane,  to  which  it  returns, 
and  from  which  it  rebounds  till  its  motion  is  destroyed.  What 
's  the  whole  space  described  by  the  body  ? 


CHAPTER  IV. 


PROJECTILES. 


In  the  preceding  chapter  we  have  discussed  the  motion  of 
a  body  by  the  joint  action  of  a  projectile  force  and  the  force 
of  gravity,  when  these  forces  were  coincident  or  opposite  in 
direction.  We  now  proceed  to  determine  the  circumstances 
of  the  motion  of  a  body,  when  the  direction  of  the  projectile 
force  is  other  than  vertically  upward  or  downward,  supposing, 
as  heretofore,  no  resistance  from  the  air. 

266.  Prop.  The  path  of  a  body,  moving  under  the  joint  in- 
fluence of  a  projectile  force,  and  the  force  of  gravity  considered 
as  a  constant,  accelerating  force,  is  a  parabola. 

Let  v  be  the  velocity  of  pro- 
jection from  the  point  A,  in  the 
direction  ANY,  and  t  the  time 
in  which  the  body  will  describe 
AN=y,  with  the  uniform  veloci- 
ty v,  if  gravity  do  not  act.  Let 
AM=a;  be  the  space  through 
which  the  body  will  fall  in  the 
time  t  by  the  action  of  gravity. 
Completing  the  parallelogram 
MN,  the  actual  place  of  the 
body  at  the  end  of  the  time  t  is  ^' 
P. 

By  (37),  AN=y=vt;  and  by 
(54),AM=NP=x=%*2.   Elim-  X 
inating  t  from  these  two  equa- 
tions, we  get 


y  — — x. 

g 


(«) 


If  h  be  the  space  due  to  the  velocity  v,  or  the  space  through 


1G2 


DYNAMICS. 


which  the  body  must  fall  to  acquire  the  velocity  of  projection 
(55),  vi—2gh,  which,  substituted  in  (a),  gives 

y'=4hx,  (62) 

the  equation  of  a  parabola  referred  to  the  oblique  axes  AX, 
AY ;  and,  since  4h  is  the  parameter  to  the  diameter  through 
A,  h  is  the  distance  from  A,  the  point  of  projection  to  the  focus 
F,  or  to  the  directrix. 

Cor.  The  velocity  of  the  body  at  any  point  of  its  path  is 
that  which  the  body  would  acquire  in  falling  vertically  from 
the  directrix  to  that  point. 

For  if  the  body  were  projected  from  any  point  of  its  path  in 
the  direction,  and  with  the  velocity  it  has  at  that  point,  it  would 
obviously  describe  the  same  path.  Therefore  the  velocity  at 
that  point  must  be  equal  to  that  due  to  one  fourth  the  parame- 
ter to  the  diameter  at  that  point,  which  is  the  distance  from 
that  point  to  the  directrix  or  to  the  focus. 

267.  Prop.  To  find  the  equation  to  the  path  of  a  projectile 
when  referred  to  horizontal  and  vertical  co-ordinate  axes. 

Let  v  be  the  velocity  of  pro- 
jection in  the  direction  AN, 
which  makes  with  AX  the 
angle  of  elevation  NAX=a, 
APB  the  path  of  the  body, 
AM=z,  PM=?/,  the  co-or- 
dinates of  the  point  P,  t  the 
time  in  which  the  body  de- 
scribes the  arc  AP,  and  pro- 
duce MP  to  meet  the  direc- 
tion of  projection  in  N. 
Then  AN  =vt,  NP=%*9,  and  MN=vt  sin.  a. 
Also,  AM.=x—vt.  cos.  a,  (a) 

and  ~PM.=y=vt  sin.  a— \gf.  (&) 

Eliminating  t  from  (a)  and  (b),  we  obtain 

v=x  tan.  a— x"1—* p-,  (63) 

y  2v  cos.  a 

and,  substituting  the  value  of  v*=2gh, 


PROJECTILES.  163 

^Xian-a-ih^ra  (64) 

268.  Def.  The  horizontal  range  of  a  projectile  is  the  distance 
AB  from  the  point  of  projection  to  the  point  where  it  strikes 
the  horizontal  plane  in  its  descent.  The  time  of  Jlight  is  the 
time  occupied  in  describing  APB.  The  height  through  which 
the  body  must  fall  to  acquire  the  velocity  of  projection  is  call- 
ed the  impetus. 

269.  Prop.  To  find  the  time  of  Jlight  of  a  projectile  on  a  hori- 
zontal plane. 

At  the  points  A  and  B  the  ordinate  y=0.     This  value  of  y, 

substituted  in  (b),  Art.  267,  gives 

vt  sin.  a—  hg?=0. 

2v  sin.  a 
.'.  t=0,  and  t= .  (65) 

The  former  value  of  t  applies  to  the  point  A,  and  the  latter 

to  the  point  B,  which  is,  therefore,  the  time  of  flight  required. 

Or,  since  v  sin.  a  is  the  vertical  component  of  the  velocity,  or 

the  velocity  of  projection  estimated  vertically,  if  this  value  of 

the  velocity  be  substituted  for  u,  in  (60),  and  we  make  s=0, 

we  get,  as  before, 

0=vt  sin.  a— ^gt7, 

.       2v  sin.  a 

t  =0,  and  t= . 

g 

270.  Prop.  To  find  the  range  of  a  projectile  on  a  horizontal 
plane. 

Put  y=0  in  (64),  and  we  have 

x* 

0—x  tan.  a— —. : —  ; 

4/i  cos.  a 

from  which  we  obtain 

for  the  point  A,  x=0, 

for  the  point  B,  x=4h  sin.  a  cos.  a, 

or  AB=R=2A  sin.  2a,  (66) 

the  horizontal  range  required. 

Cor.  1.  The  horizontal  range  is  greatest  when  a=45°. 

For  in  this  case      2a  =  00°,  and  sin.  2a=l.  .•.  x=2h, 


1G4  DYNAMICS. 

or  the  greatest  horizontal  range  is  equal  to  twice  the  height 
due  to  the  velocity  of  projection,  or  twice  the  distance  from 
the  point  of  projection  to  the  focus  of  the  trajectory. 

Cor.  2.  The  range  is  the  same  for  any  two  angles  of  eleva- 
tion, the  difference  between  which  and  45°  is  the  same,  or  for 
(45±0). 

For  sin.  (90° +26)  =  sin.  (90° -26), 

or  sin.  2(45°+ 6)= sin.  2(45°- 6). 

If,  therefore,  we  put  either  45°+0  or  45°  —  6  for  a  in  (66), 

the  value  of  R  remains  the 
/  same.     Thus,  if  AB  be  the 

range  of  a  projectile  when 
-p    \  the  angle  of  elevation  is  45°, 

and  APB  its  path,  the  range 
AB'  will  be  that  due  to  the 
elevation  45°— 6,  for  which 
B'  B     the  path  is  APB',  and  to 

45°  +  6,  for  which  the  path  is  AP"B'. 

Cor.  3.  When  the  velocity  of  projection  is  given,  and  we 
know  the  range  R  due  to  the  elevation  a,  we  can  readily  find 
the  range  R'  due  to  any  other  elevation  a'. 

For  R=2/*sin.  2a, 

and  R'=2/i  sin.  2a'. 

_      sin.  2a'  ,     k 

•••  R'=inr^-R-  <67> 

Cor.  4.  Since  the  horizontal  velocity  of  a  projectile  is  urn- 
form,  the  range  is  equal  to  the  horizontal  component  of  the  ve- 
locity into  the  time  of  flight,  or 

_  2v  sin.  a    2v*  sin.  a.  cos.  a. 

K=v.  cos.  aX = =2h  sin.  2a,  as  before. 

g  g 

271.  Prop.  To  find  the  greatest  height  which  a  projectile  at- 
tains. 

The  greatest  height  H  is  evidently  the  value  of  the  ordinate 

at  the  middle  of  AB,  or  when  the  time  is  one  half  the  time  of 

„.  .        -^  v  sin.  a .    ,,. 

flight.     Putting  t= in  (b),  Art.  267,  we  have 


PROJECTILES. 


165 


H= 


v  sin.  a 
g 


v  sin.  a 


tv  sin.  a 


g 
h  sin.2  a. 


(68) 


272.  Prop.  To  find  the  co-ordinates  of  the  point  where  a  pro- 
jectile will  strike  an  inclined  plane  passing  through  the  point 
of  projection,  the  range  jn  the  inclined  plane,  and  the  time  of 
flight. 

Let  y=x  tan.  (3  be  the  equation 
of  the  line  AC,  which  is  the  in- 
tersection of  the  inclined  plane, 
with  the  vertical  plane  of  the 
path  of  the  body. 

Substituting  this  value  of  y  in 
(64),  we  obtain,  after  reduction, 
for  the  abscissa  of  the  point  C, 


M    33 


_4h  cos.  a.  sin. (a— (3) 
cos.  (3 


(a) 


By  substituting  in  (64)  for  x,  its  value  y  cot.  (3,  and  reducing, 
we  get  for  the  ordinate  of  C, 

4h.  cos.  a.  sin.  (3.  sin.  (a— (3) 


y~  cos.2/3 

To  find  AC=R',  multiply  (a)  by  sec.  (3= 


cos.  j3' 


(*) 


and   we 


have 


kri    „.  _    4/j.  cos.  a.  sin.  (a— 0)       4     . 

AC=R'=a;sec.|3= — -i ^.     (G9) 

cos.  (3  v     7 

If  the  inclined  plane  cut  the  path  of  the  projectile  below  the 
axis  AX,  (3  will  be  negative. 

The  time  of  flight  is  equal  to  the  time  of  describing  the  ab- 
scissa AM  with  the  horizontal  component  of  the  velocity. 
Hence  (37),  dividing  (a)  by  v  cos.  a.,  we  get 

4/tsin.  (a— (3) 


v.  cos.  (3     ' 
2vs'm.(a—(3) 
g.  cos.  (3 


(70) 


166  DYNAMICS. 

273.  Schol.  It  has  been  found  by  experiment,  that  if  w  be 
the  weight  of  a  ball  or  shell,  p  the  weight  of  the  gunpowder 
used  in  discharging  the  ball  or  shell  from  a  mortar,  and  v  the 
velocity  generated  by  the  powder, 


\/^  feet.  (71) 

V    w 


u=1600 

w 


274.    EXAMPLES. 

Ex.  1.  A  body  is  projected  at  an  angle  of  elevation  of  15° 
with  a  velocity  of  60  feet.  Find  the  horizontal  range,  the 
greatest  altitude,  and  time  of  flight. 

,      x    ,      u3     3600 
From  (54'),  h=— =-—^-=55.96. 

2g       64g- 

From  (66),  R=2A  sm.2a=2h.  sin.  30°=2./L|=/*=55.96 
From  (68),  H  =  A  sin.9  a.  Log.  sin.  a=  9.4129962 

"       "      =  9.4129962 

h  =  1.7478777 


H  =  3.7486 "  H  =  .0.5738701 


„  _     2v  sin.  a 

From  (65),  T= .  Log.  2v=  2.0791812 

Log.  sin.  a=  9.4129962 

a.c.     "  g=  8.4925940 

.-.   T  =  0.96554 "  T=  9.9847714 

Ex.  2.  A  body  is  projected  at  an  angle  of  elevation  of  45° 
and  descends  to  the  horizon  at  a  distance  of  500  feet  from  the 
point  of  projection.  Required  the  velocity  of  projection,  the 
greatest  altitude,  and  time  of  flight. 

Ans.  v  =126.82  feet 
H=125. 

T=5.58  seconds. 
Ex.  3.  The  horizontal  range  of  a  projectile  is  1000  feet  and 
the  time  of  flight  is  15  seconds.     Required  the  angle  of  eleva- 
tion, velocity  of  projection,  and  greatest  altitude. 

Ans.  a  =74o.33'.09". 
v  =250.29  feet. 
H=904.69     " 


PROJECTILES.  167 

Ex.  4.  If  a  body  be  projected  at  an  angle  of  elevation  of 
60°,  with  a  velocity  of  850  feet,  find  the  parameter  to  the 
axis  of  the  parabola  described,  and  the  co-ordinates  of  the 
focus. 

Ans.  p=  11230.57  feet. 
x=  9725.67     " 
y=  5615.28     " 

Ex.  5.  Find  the  velocity  and  angle  of  elevation  of  a  ball  that 
it  may  be  100  feet  above  the  ground  at  the  distance  of  one 
quarter  of  a  mile,  and  may  strike  the  ground  at  the  distance 
of  one  mile. 

Ans.  a=5°.46'.04".6. 
v=921. 566  feet. 

Ex.  6.  What  must  be  the  angle  of  elevation  of  a  body  in 
order  that  the  horizontal  range  may  be  equal  to  three  times  the 
greatest  altitude  ?  What,  that  the  range  may  be  equal  to  the 
altitude? 

Ex.  7.  A  body  is  projected  at  an  angle  of  elevation  of  60°, 
with  a  velocity  of  150  feet.  Find  the  co-ordinates  of  its  posi- 
tion, its  direction,  and  velocity  at  the  end  of  5  seconds. 

Ex.  8.  A  body  is  projected  from  the  top  of  a  tower  200  feet 
high,  at  an  angle  of  elevation  of  60°,  with  a  velocity  of  50  feet. 
Find  the  range  on  the  horizontal  plane  passing  through  the 
foot  of  the  tower,  and  the  time  of  flight. 

Ex.  9.  A  body,  projected  in  a  direction  making  an  angle  of 
30°  with  a  plane  whose  inclination  to  the  horizon  is  45°,  fell 
upon  the  plane  at  the  distance  of  250  feet  from  the  point  of  pro- 
jection, which  is  also  in  the  inclined  plane.  Required  the  ve- 
locity of  projection,  and  the  time  of  flight. 

Ex.  10.  The  heights  of  the  ridge  and  eaves  of  a  house  are 
40  feet  and  32  feet  respectively,  and  the  roof  is  inclined  at  30° 
to  the  horizon.  Find  where  a  sphere  rolling  down  the  roof 
from  the  ridge  will  strike  the  ground,  and  also  the  time  of  de- 
scent from  the  eaves. 

Ex.  11.  How  much  powder  will  throw  an  eight-inch  shell, 
weighing  48  lbs.,  1500  yards  on  an  inclined  plane,  the  angle  of 


168  DYNAMICS. 

elevation  of  the  plane  being  28°.45',  and  that  of  the  mortar  be- 
ing 48°.30'  ? 

Ex.  12.  Find  the  velocity  and  angle  of  elevation  that  a  pro 
jectile  may  pass  through  two  points  whose  co-ordinates  are 
x=300  feet,  ^=60  feet,  z'=400  feet,  and  y'=40  feet.  Also 
find  the  horizontal  range,  greatest  altitude,  and  time  of  flight. 


CHAPTER  V. 


CONSTRAINED     MOTION. 
§  I.    MOTION    ON    INCLINED    PLANES. 

275.  Prop.  To  determine  the  relations  of  the  space,  time,  and 
velocity  when  a  body  descends  by  the  action  of  gravity  down  an 
inclined  plane. 

Let  the  body  fall  from  C  down 
the  inclined  plane  CA,  whose  in- 
clination is  a,  M  be  the  position  of 
the  body  at  any  time  t  from  rest  at 
C,  CM=s,  and  v—  the  velocity  at 
M. 

The  force  of  gravity  g,  acting  in  the  direction  Mg,  may  be 
resolved  into  two  forces,  one  f=g  sin.  a  acting  in  the  direction 
CA,  the  other  g  cos.  a  acting  perpendicularly  to  CA,  and  wholly 
ineffective  in  producing  motion ;  the  body  is,  therefore,  urged 
down  the  inclined  plane  by  the  constant  accelerating  force 
f=g.  sin.  a. 

If,  therefore,  this  value  of/  be  substituted 
in  (53),  we  have  v  —gt  sin.  a,  (72) 

in  (54),  we  have  s  =%gf  sin.  a,  (73) 

in  (55),  we  have  v"—2gs  sin.  a,  (74) 

from  which  all  the  circumstances  of  the  motion  may  be  de- 
termined. 

276.  Prop.  The  velocity  acquired  by  a  body  in  falling  down 
an  inclined  plane  is  equal  to  that  acquired  in  falling  freely 
through  the  height  of  the  plane. 

If  s=AC,  the  length  of  the  plane,  and  /i=BC,  the  height,  by 
(74), 

v'!=2g.s.  sin.  a 
=2g.CB. 


170 


DYNAMICS 


.*.  v  =  V2ght 

which,  by  (53'),  is  the  velocity  due  to  h,  the  height  of  the 
plane. 

277.  Prop.  The  times  of  descent  down  different  inclined  planes 
of  the  same  height  vary  as  the  lengths  of  the  planes. 

By  (73),  s=k'5  sin-  «> 

BC 


or 


AC=%«' 


••  <=ACV; 


2 

^:bc 

x  AC,  when  BC  is  constant. 

278.  Prop.  To  find  the  relations  of  the  space,  time,  and  veloc- 
ity when  a  body  is  projected  down  or  up  an  inclined  plane. 

Substituting  for  f,  its  value,  g  sin.  a,  on  an  inclined  plane, 
in  (59),  we  have  v  =vtdtgt  sin.  a,  (75) 

in  (GO),  we  have  s  =vlt±^gt'1  sin.  a,  (76) 

in  (61),  we  have         v*=v\±2gs  sin.  a,  (77) 

which  give  ail  the  circumstances  of  the  motion  of  the  body. 

279.  Prop.  The  times  of  descent  down  all  the  chords  of  a  cir- 
cle in  a  vertical  plane,  drawn  from  either  extremity  of  a  vertical 
diameter,  are  the  same,  and  equal  to  that  down  the  vertical  di- 
ameter. 

Let  AB  be  the  vertical  diameter,  BD 
and  AD  any  chords  drawn  from  its  ex- 
tremities, and  DC  perpendicular  to  AB. 
To  find  the  time  of  descent  down  BD 
we  have,  from  (73), 

BD: 

„BC 


■  \gf  sin.  B DC 


,_2BD2 
~>.BC 

2.BA 


8 


CONSTRAINED     MOTION. 


17] 


or 


t- 


which,  by  (5C),  equals  the  time  down  the  diameter 
In  the  same  manner, 

DA=fcfCj 


DA' 


from  which  we  obtain,  as  before, 


2.BA 


g 

280.  Prop.   To  find  the  straight  line  of  quickest  descent  from 
a  given  point  within  a  vertical  circle  to  its  circumference. 

Let  P  be  the  given  point.  Draw  the 
vertical  radius  CA,  join  AP  and  produce 
it  to  meet  the  circumference  at  B ;  PB 
will  be  the  line  required. 

Join  BC  and  draw  POD  parallel  to 
AC.  Since  BC=AC,  BO=PO.  With 
O  as  a  center,  and  radius  PO,  describe 
the  circle  PBD,  which  will  be  tangent  to 
the  circle  C  at  B. 

Now  the  time  down  PB  will  equal  the  time  down  the  verti 
cal  diameter  PD,  and  the  time  down  every  other  chord  drawn 
from  P  {Art.  279).  But  any  other  chord  from  P,  produced  to 
the  circumference  of  C,  will  be  partly  without  the  circum- 
ference of  O,  and,  therefore,  the  time  down  it  will  be  greater 
than  the  time  down  PB,  which  is  therefore  the  line  of  quickest 
descent  to  the  circumference  of  C. 

281.  Prop.   To  find  the  straight  line  of  quickest  descent  from 
a  given  point  to  a  given  inclined  plane. 

Let  A  be  the  given  point,  AG  a  vertical 
line  passing  through  this  point,  and  AB  a 
perpendicular  to  the  inclined  plane  BG. 
The  line  AC,  bisecting  the  angle  BAG, 
will  be  the  line  required. 

Draw  CE  parallel  to  BA,  and  therefore 
perpendicular  to  BG.     Since  EAC=CAB 


Q 


172  DYNAMICS. 

=  ECA,  the  triangle  EAC  is  isosceles.  With  E  as  a  centei, 
and  EA  or  EC  as  a  radius,  describe  the  circle  ACD,  to  which 
BCG  will  be  a  tangent. 

Now  the  time  down  i\.C  will  be  equal  to  the  time  down  AD, 
and  to  the  time  down  every  other  chord  drawn  from  A.  But 
any  other  chord  of  the  circle  must  be  produced  to  meet  the 
plane  BG.  Therefore  the  time  down  any  other  line,  drawn 
from  A  to  the  plane,  will  be  greater  than  the  time  down  AC, 
which  must  be  the  line  of  quickest  descent  required. 

282.  Prop.  Two  bodies  are  suspended  from  the  extremities  oj 
a  cord  passing  over  a  fixed  pulley :  to  determine  the  circumstances 
of  their  motion. 

a  Let  P  and  Q,  be  the  weights  of  the  bodies,  of  which 

P        Q 

P  is  the  greater.     By  (22)  their  masses  are  —  and  — 
&  j  g  g 

respectively.  Neglecting  the  rigidity  of  the  cord, 
the  inertia  and  friction  of  the  pulley,  if  P=Q,  they 
will  counterpoise,  and  no  motion  will  ensue  ;  but 
P>Q,  P  will  descend,  and  Q,  rise  through  equal 
spaces  by  a  force  equal  to  the  difference  P— Q,  of 
their  weights.  But  this  being  a  moving  force  {Art. 
225),  is  equal  to  the  accelerating  force  into  the  mass 

moved,   =/( ) ,  where  /  represents    the    accelerating 

force. 

Hence  p_Q=/?±5 

g 

,     P-Q 

This  being  a  constant  accelerating  force,  by  substituting  its 
value  in  (53),  (54),  and  (55),  we  have  expressions  for  the 
space,  time,  and  velocity. 

283.  Schol.  If  the  inertia  I  of  the  pulley  be  taken  into  con- 
sideration, \g  is  an  additional  force  to  be  overcome  by  the 
difference  of  the  weights,  and  in  this  case 

P-Q 


® 


© 


f=g-v 


P+Q+Ig-" 


CONSTRAINED     MOTION.  173 

This  is  the  formula  for  Atwood's  machine,  an  instrument  fo* 
illustrating  the  laws  of  falling  bodies.  The  friction  of  the  pul- 
ley is  reduced  by  friction  wheels,  and  the  rigidity  of  the  cord 
by  employing  a  fine  flexible  thread.  By  making  the  difference 
between  P  and  Q,  small,  the  motion  is  made  so  slow  as  to  ren- 
der the  time,  space,  and  velocity  easily  determinable. 

§  II.    MOTION    IN    CIRCULAR    ARCS. 

284.  Prop.  When  a  body  descends  by  the  action  of  gravity 
down  any  smooth  arc  of  a  circle  in  a  vertical  plane,  the  velocity 
at  the  lowest  point  is  proportional  to  the  length  of  the  chord  of 
the  arc. 

Let  a  body  descend  from  the  point  D  {Fig.,  Art.  279)  down 
the  arc  DA.  Since  the  reaction  of  the  curve  is  always  per- 
pendicular to  the  path  of  the  body,  it  can  neither  accelerate 
nor  retard  the  motion  of  the  body.  Its  velocity,  therefore,  at 
A  will  be  that  which  it  would  acquire  in  falling  through  the 
vertical  height  C  A  of  the  arc. 
Hence,  by  (55),  ua=2g€A 

ADS 
=2*AB_ 

ocAD. 

When  the  arcs  are  small,  the  velocities  are  nearly  proportional 
to  the  arcs,  the  principle  referred  to  in  Art.  254. 

285.  Prop.  When  a  body  is  constrained  to  describe  the  sides 
of  a  polygon  successively,  to  find  the  velocity  lost  in  passing  from 
one  side  to  the  succeeding  one. 

Let  AB  and  BD  be  two  adjacent  sides 
of  the  polygon,  w  the  angle  made  by  their 
directions,  and  v  the  velocity  of  the  body 
at  the  point  B.  In  passing  from  AB  to 
BD  some  of  the  velocity  will  necessarily  be  lost.  Resolving 
the  velocity  in  the  direction  BD,  we  have,  for  the  velocity  in 
BD, 


174 


DYNAMICS. 


V  COS.  W. 

This,  subtracted  from  the  primitive  velocity  v,  will  give  for  the 
velocity  lost, 

Uj=U  —  v  cos.  w 
=v(l  —  cos.  w) 
=  v  versin.  w 
v  sin.2  w 


2  — versin.  w 

286.  Cor.  If  the  sides  of  the  polygon  be  increased  in  num 
ber,  the  angle  w  diminishes  ;  and  when  the  polygon  becomes  a 
circle,  iv,  and  consequently  its  sine,  becomes  indefinitely  small. 
In  this  case,  versin.  w,  small  in  comparison  with  2,  may  be  re- 
iected,  and 

v    ■    a 
v ,  =-  sin.  w. 

But  if  sin.  w  is  infinitely  small,  sin.2u>  is  infinitely  smaller,  and 
the  velocity  lost 

When,  therefore,  a  point  or  body  is  constrained  to  describe  a 
curve,  no  velocity  is  lost  by  the  reaction  of  the  curve. 

287.  Prop.  If  a  material  point  move  through  one  side  of  a 
regular  polygon  with  a  uniform  velocity,  to  find  the  direction 
and  intensity  of  the  impulse  which  must  be  given  to  the  material 
■point  at  each  angle  of  the  polygon  in  order  that  it  may  describe 
the  entire  polygon  with  the  same  uniform  velocity. 

Let  ABC  ...  be  the  polygon,  and  let 
D  the  material  point  describe  AB  with  the 
velocity  v  in  the  unit  of  time.  If,  when 
\z  at  B,  no  other  force  act  upon  the  point, 
it  will  describe  BD=AB  in  the  same 
time.  Join  DC,  and  draw  BN  equal 
and  parallel  to  DC.  If,  when  at  B,  the 
point  receive  an  impulse  in  the  direc- 
tion BN,  such  as  to  cause  it  to  describe 
BN  in  the  same  time  that  it  would  describe  BD,  the  point  will 
describe  the  diagonal  BC,  the  succeeding  side  of  the  polygon. 


CONSTRAINED      MOTION.  175 

in  the  same  time.  But  since  BC  =  BD,  the  velocity  in  BC  is  the 
same  as  that  in  AB,  and  the  point  will  describe  the  two  adjacent 
sides  with  the  same  uniform  velocity.  Now,  since  the  trian- 
gle BCD  is  isosceles,  BDC=BCD=CBN.  But  ABN=BDC  ; 
.-.  ABN=CBN,  and  BN  bisects  the  angle  ABC.  Hence  the 
direction  of  the  impulse  will  pass  through  the  center  of  the  cir- 
cumscribing circle,  and,  if  similar  impulses  be  applied  at  each 
of  the  angles  of  the  polygon,  the  point  will  describe  the  entire 
polygon,  with  the  original  velocity  unchanged. 

To  find  the  magnitude  of  the  component  force  BN=f,  let  r 
be  the  radius  of  the  circle ;  and,  since  BN  is  perpendicular  to 
the  chord  AC, 

2r 

RC2 
and  /=BN= . 

But  BC  =  AB  is  the  space  described  in  the  unit  of  time,  and  is 
therefore  represented  by  v.     Hence 

or,  the  intensity  of  the  impulse  is  equal  to  the  square  of  the  ve- 
locity in  the  polygon  divided  by  the  radius  of  the  circumscribing 
circle. 

288.  Cor.  The  force  which  must  continually  urge  a  material 
point  toward  the  center  of  a  circle,  in  order  that  it  may  de- 
scribe the  circumference  with  a  uniform  velocity,  is  equal  to 
the  square  of  the  velocity  divided  by  the  radius. 

Since  the  reasoning  in  the  proposition  is  independent  of  the 
number  of  sides  in  the  polygon,  the  sides  of  the  polygon  may 
be  increased  in  number,  and  the  frequency  of  the  impulse  in- 
creased in  the  same  ratio,  without  affecting  the  relation  of  the 
intensity  of  the  impulse  to  the  velocity  in  the  polygon.  But 
when  the  number  of  sides  becomes  infinite,  or  the  polygon  be- 
comes a  circle,  the  impulses  will  no  longer  be  successive,  but 
an  incessant  action  of  the  same  intensity. 

The  force  is,  therefore,  a  constant  accelerating  force,  and 


176  DYNAMICS. 


V 

If  the  mass  of  the  body  be  taken  into  consideration,  the 
moving  force  {Art.  225)  will  be 

F=mf=^.  (80) 

It  is  obviously  immaterial  whether  the  body  be  retained  in 
ts  path  by  the  reaction  of  a  smooth  curve,  or  by  an  inextens- 
ible  cord  without  weight,  by  which  it  is  connected  with  the 
center  of  the  circle.  F  will  be  a  measure  of  the  resistance  of 
the  curve  in  the  one  case,  and  of  the  tension  of  the  cord  in  the 
other. 

289.  Def.  The  force  which  constantly  urges  a  body  toward 
the  center  of  its  circular  path  is  called  a  centripetal  force.  The 
tendency  which  the  body  has  to  recede  from  the  center,  in  con- 
sequence of  its  inertia,  or  the  resistance  which  it  offers  to  a  de- 
flection from  a  rectilinear  path,  the  resistance  being  estimated 
in  the  direction  of  the  radius,  is  called  a  centrifugal  force. 

290.  Prop.  To  discuss  the  circumstances  of  the  motion  of  a 
body  constrained  to  move  in  a  circle  by  the  action  of  a  central 
force. 

1°.  When  the  masses  are  equal,  by  (79), 

j     r 

Tf  r  be  constant, 

/  ac  v- ; 

or,  when  a  body  moves  in  the  circumference  of  a  circle  by 
means  of  a  cord  fixed  at  the  center,  the  tension  of  the  cord,  ot 
the  centrifugal  force,  will  vary  as  the  square  of  the  velocity. 
2°.  If  v  be  constant, 

/4 

r 

or,  if  equal  bodies  describe  different  circles  with  the  same  ve- 
locity, the  centrifugal  forces  will  be  inversely  as  the  radii  of  the 
circles. 


or 


CONSTRAINED     MOTION.  177 

3°.  Let  T  :e  the  periodic  time,  or  time  of  one  revolution. 
Then  (37)  vT==2nrt 

,     4ttV2 

(79),  =fr. 

4?rV 
••■  f=-Tr.  (81) 

r 

or,  the  centrifugal  force  varies  directly  as  the  radius  of  the  circle, 
and  inversely  as  the  square  of  the  periodic  time. 
4°.  If  T  ac  r\  by  substitution  in  (81), 
,     r       1 

or,  when  the  squares  of  the  periodic  times  are  as  the  cubes  oi 
the  distances  from  the  center,  the  centrifugal  force  will  be  in- 
versely as  the  square  of  the  distance. 

5°.  If  o)  be  the  angle  subtended  by  the  arc  described  by  the 
body  in  the  unit  of  time,  w  is  called  the  angular  velocity.  As- 
suming for  the  angular  unit  the  angle  whose  arc  is  equal  in 
length  to  the  radius,  v=ru.  Substituting  this  value  of  v  in 
(79), 

f=ror-  (82) 

or,  the  centrifugal  force  varies  as  the  product  of  the  radius  and 
the  square  of  the  angular  velocity. 

6°.  If  in  each  of  the  above  cases  the  masses  be  not  the  same, 
the  centrifugal  forces  will  vary  directly  as  the  masses  in  addi- 
tion to  the  other  causes  of  variation. 

The  foregoing  principles  admit  of  a  very  simple  and  satis- 
factory illustration  by  means  of  an  instrument  called  the  whirl- 
ing table. 

291.  Prop.  To  find  the  relation  of  the  centrifugal  force  in  a 
circle  to  the  force  of  gravity. 

Let  h  be  the  height  due  to  the  velocity  v  which  the  body 
has  in  the  circle.     By  (53')  vi—2gh.     This  value  of  ds,  substi 
tuted  in  (79),  gives 

M 


178  DYNAMICS. 

fJ*gh 

J-  r  > 

f     2h 
or  -= — . 

g      r 

Hence  the  centrifugal  force  is  to  the  force  of  grivily  as  twice 
the  height  due  to  the  velocity  in  the  circle  is  to  the  radius  of  the 
circle. 

Cor.  Uf=g,  r=2h, 

or,  in  a  circle  whose  radius  is  equal  to  twice  the  height  due  to 
the  velocity  in  the  circle,  the  centrifugal  force  is  equal  to  the 
force  of  gravity. 

292.  Prop.   To  find  the  centrifugal  force  at  the  equator. 

By  (8i),  /=^?- 

The  equatorial  radius  R  of  the  earth  is  3962.G  miles  =20,922,528 
feet.  7r=3.1415926.  And,  since  the  earth  revolves  on  its  axis 
in  0.997269  of  a  day,  T=0.997269X864005=861649.  These 
values,  substituted  in  the  above,  give 

/=0.1 11255  feet. 

293.  Schol.  Since  the  force  of  gravity  g  at  the  equator  has 
been  found  to  be  32.08954  feet,  if  G  be  the  force  of  gravity  on 
the  supposition  that  the  earth  does  not  revolve  on  its  axis,  then 

g=G-f 
or  G=g+f, 

=  32.08954+0.1 1 1255=32.200795, 
f      0.111255       1 
and  G=32.200795=289nearly;  (83) 

or  the  centrifugal  force  at  the   equator  is  — —  the  force  of 

289 

gravity. 

294.  Prop.  To  find  the  time  in  which  the  earth  must  revolve 
on  its  axis,  in  order  that  the  centrifugal  force  at  the  equator 
may  equal  the  force  of  gravity. 

Let  T'  be  the  required  time,  and  f  the  corresponding  cen 
trifugal  force.     From  (79)  we  have 


COXS  T  R  A  i  N  E  U     MOTIO  N. 


179 


R.     JT 
f'-f-'F'  T"' 
But  in  this  case  R=R',  and/'  is  to  become  equal  to  G. 

.-./:G= 


1       1 

"fT12    •    rp/2» 


or 


and 


.J  rp2 


rp2 TH 

G     ~289     ' 

T'  =— f  nearly. 

17  J 

1 


Hence  the  earth  must  revolve  in  y-th  of  its  present  period. 

295.  Prop.  The  centrifugal  force  diminishes  gravity  at  dif- 
ferent places  on  the  earth's  surface  in  the  ratio  of  the  square  of 
the  cosine  of  the  latitude. 

Regarding  the  earth  as  a  sphere,  from 
which  it  differs  but  very  little,  let  EF  be 
the  axis,  AC  =  R  the  equatorial  radius,  B 
any  point  whose  latitude  is  the  arc  AB, 
measured  by  the  angle  ACB=0,  and 
BG=R'  the  radius  of  the  parallel  of  lati- 
tude passing  through  B.  Let  the  centrif- 
ugal force  at  B,  which  acts  in  the  direc- 
tion GB,  be  represented  by  BD. 

47T2R' 

By  (81),  BD^4^. 

Resolving  this  in  the  direction  of  the  vertical  CZ,  opposite  to 
that  in  which  gravity  acts,  and  calling  this  component/',  we 
have 


/'  =  B6=-7pj-cos.0. 


But  R'==R  cos.  (p, 


■'.  f- 


4tt!R 


COS.*  0, 


T,    ™  r,  (84) 

in  which  the  coefficient  of  cos.2  0  is  constant  for  all  latitudes. 
Hence  f  ac  cos.2  0, 


180  DYNAMICS. 

f  being  the  diminution  of  gravity  by  the  actio.Q  of  the  centrif- 
ugal force. 

The  latitude  of  Middletown  being  41°.33'.10",  the  centrif- 
ugal force/',  in  a  vertical  direction  at  this  place,  will  be  found 
to  be 

f'= 0.062305. 

If  this  be  added  to  the  observed  gravity  g'=32. 16208,  we  shall 
nave  for  the  whole  gravity,  undiminished  by  the  centrifugal 
force, 

•    G' =£'+/'= 32.224385. 

296.  Cor.  Resolving  the  centrifugal  force  at  B  in  a  direc 
tion  perpendicular  to  the  radius  CB,  we  have 

4ttR 
f'=Bd=-~r  cos.  <(>  sin.  <p, 

2tt2R   . 
=-7jii-  sin.  2(f). 

If,  therefore,  the  matter  of  the  earth  were  susceptible  ot 
yielding  to  this  component  of  the  centrifugal  force,  it  would 
necessarily  cause  the  earth  to  deviate  from  a  spherical  form. 
The  fluid  portions  of  the  surface  are  therefore  urged  toward 
the  equatorial  regions,  thereby  increasing  the  equatorial  diame- 
ter and  diminishing  the  polar.  If  the  solid  portion  did  not  also 
partake  of  the  same  general  form  as  the  fluid,  we  should  ex- 
pect to  find  vast  equatorial  oceans  and  polar  continents,  with 
polar  mountains  far  exceeding  the  equatorial  in  height.  But 
the  actual  distribution  of  the  waters  and  mountains  on  the  sur- 
face of  the  earth  is  widely  different.  The  amount  of  deviation 
from  a  spherical  form,  which  the  earth  must  take  from  the  ac 
tion  of  the  forces  developed  by  its  motion,  will  depend  in  part 
on  the  law  of  variation  of  its  density  from  the  surface  to  the 
center.  By  measurement  the  equatorial  diameter  is  found  to 
exceed  the  polar  about  26  miles.  The  ratio  of  the  difference 
of  the  equatorial  and  polar  radii  to  the  equatorial  radius,  called 

the  compression  or  ellipticity  of  the  earth,  is  about  — — . 

297.  Prop.  To  find  'he  centrifugal  force  of  the  moon  ir.  its 
orbit. 


CONSTRAINED     MOTION.  181 

Let  R  be  the  radius  of  the  earth,  nR  the  mean  radius  of  the 
moon's  orbit,  and  P  the  periodic  time  of  the  moon  in  seconds. 
Then  the  velocity  of  the  moon  will  be 

2miR 
»=— - 

This,  substituted  in  (79),  gives,  for  the  centrifugal  force, 

„    47T2nR 
/=-pi--  («) 

If  the  earth  were  an  exact  homogeneous  sphere,  the  intensity 
of  gravity  would  be  the  same  at  all  points  on  the  surface.  Be- 
ing a  spheroid,  differing  little  from  a  sphere,  the  theory  of  at- 
traction of  spheroids  shows  that  it  is  necessary  to  use  the  in- 
tensity of  gravity  at  a  latitude  <*,.  of  which  the  sine  is  the  V^ ; 
whence  0=35°. 15'. 52".  Supposing  the  compression  of  the 
earth  to  be  0.00324,  in  the  latitude  <f>, 

R=20897947  feet. 

To  find  n,  conceive  two  lines  drawn  from  the  center  of  the 
moon,  one  to  the  center  of  the  earth,  and  the  other  tangent  to 
the  earth  at  the  equator.  The  angle  made  by  these  lines  at 
the  center  of  the  moon,  and  subtended  by  the  equatorial  radius 
of  the  earth,  called  the  moon's  equatorial  horizontal  parallax, 
is  found  by  astronomical  observations  to  be  at  a  mean  57'.  1". 
This  angle,  when  subtended  by  the  radius  of  the  earth  in  the 

,     ,  .  R  i 

latitude  of  35°.15'.52",  is  7r=56'.57".33.     Now  -5-=  sin.  *=-, 

from  which  we  find  ?i=60.3612.  P=2360585\     These  values, 

substituted  in  (a),  give 

J=0.00894. 

298.  Cor.  The  moon  is  retained  in  its  orbit  by  the  gravity 

of  the  earth.     For,  since  the  intensity  of  gravity  is  inversely  as 

the  square  of  the  distance  from  the  earth's  center,  its  intensity 

g'  at  the  moon  may  be  found  by  the  proportion 

_  1         1 

g:g'~}V:'rfR*; 

g 
whence  g'      =-5. 

The  intensity  of  gravity  in  the  latitude  of  35°.15'.52",  uti 


182 


DYNAMICS. 


diminished  by  the  centrifugal  force  of  the  earth  in  its  diurna. 
motion,  is 

£•=32.24538. 

Substituting  for  n  and  g  their  values,  we  find          g 
g'  =  0.00885. 

The  difference  between/  and  g'  is  less  than  one  ten  thousandth 
of  a  foot,  a  difference  which  may  be  attributed  to  errors  of  ob- 
servation. 


§  III.    PENDULUM. 

299.  Def.  A  simple  pendulum  is  a  material  point  suspended 
by  a  right  line,  void  of  weight,  and  oscillating  about  a  fixed 
point  by  the  force  of  gravity.  The  path  of  the  point  is  the  arc 
of  a  vertical  circle,  of  which  the  fixed  extremity  of  the  line  is 
the  center. 

300.  Pkop.  To  find  the  force  by  which  the  pendulum  is  urged 
in  the  direction  of  its  path. 

Let  M  be  the  material  point  suspend- 
ed from  C  by  the  line  CM.  When  re- 
moved from  its  vertical  direction  CM 
to  the  inclined  position  CM',  gravity 
will  cause  the  point  to  descend  and  de- 
scribe the  arc  M'M.  Resolving  the 
force  of  gravity  g  in  the  directions 
M'B  and  M'T,  parallel  and  perpen- 
dicular to  the  radius  CM',  the  com- 
ponent g  cos.  6,  in  the  direction  of  CM', 
will  be  counteracted  by  the  fixed  point  C.  The  component  g 
sin.  0,  in  the  direction  of  the  tangent  M'T,  will  be  the  only 
effective  force  to  produce  motion.  Hence  the  accelerative 
force  is 

f=g  sin.  6. 

Let  CM.=l,  the  length  of  the  pendulum  and  the  arc  M'M=* 
the  semi-arc  of  the  vibration;  then  s=W. 

.-.  f=g.  sin.  -. 


CONSTRAINED     MOTION.  183 

Now  if  x  be  any  arc  of  a  circle,  it  is  shown  in  trigonometry 
that 

x» 

Hence  ^(j_^+_±__),  &c. 

But  when  the  arc  5  is  small  compared  with  I,  the  cube  and 
higher  powers  of  j  may  be  neglected,  and 

cz  s, 
or  the  force  varies  as  the   distance    from  the  lowest  point 
measured  on  the  arc. 

301.  Prop.  When  a  body  is  urged  toward  a  fixed  point  by  a 
force  varying  directly  as  its  distance  from  that  point,  the  times 
of  descent  to  that  point  from  all  distances  will  be  the  same. 

For  the  space  described  will  obviously  depend  on  the  in- 
tensity of  the  force  and  the  time  of  its  action,  or  will  be  meas- 
ured by  the  product  of  the  force  by  some  function  of  the  time. 

Hence  s—f.(ji{i). 

Since  s  varies  as/,  let  s=nf,  n  being  the  constant  ratio  of  the 
space  to-  the  force  ;  then,  by  substitution, 

n=<f)(t)  ; 
or  the  function  of  the  time  is  the  same  whatever  be  the  dis 

tance,  and,  in  the  case  of  the  pendulum,  n=-.     (85). 

302.  Schol.  It  will  be  shown  hereafter  that  the  time  of  de 
scent  to  the  center  of  force,  or  the  point  toward  which  the 

7T     /I 
force  is  directed,  is  always  equal  to  -\/-,  when  u  is  the  ac- 

g 
celerating  force.     In  the  case  of  the  pendulum,  n=-j,  and  we 

shall  have,  for  the  time  of  descent  to  the  lowest  point,  or  the 
time  of  a  semi-vibration. 


184  DYNAMICS. 

2V  g 

When  the  material  point  has  reached  its  lowest  pos  tion,  its 
momentum  will  cause  it  to  rise  on  the  other  side  of  the  vertical 
line ;  and,  since  its  velocity  at  the  lowest  point  will  be  that  due 
to  the  vertical  height  fallen  through,  and  the  force  will  require 
the  same  time  to  destroy  the  velocity  that  was  required  to 
generate  it,  the  whole  time  of  a  vibration  will  be 


'=*Vv  <86> 


As  the  circumstances  of  the  material  point  at  the  end  of  the 
time  T  are  the  same  as  at  the  commencement,  it  will  then  per- 
form another  vibration  in  the  same  time,  and  so  continuing,  all 
its  vibrations  will  be  isochronal. 

It  should  be  recollected  that  these  results  are  obtained  on 
the  supposition  that  the  arcs  of  vibration  are  small. 

303.  Prop.   To  discuss  the  circumstances  of  the  vibration  of 
different  pendulums. 

1°.  Since  in  (86)  it  is  constant, 

TocXl; 

or,  the  time  of  vibration  of  a  pendulum  varies  directly  as  the 
square  root  of  the  length,  and  inversely  as  the  square  root  of  the 
accelerating  force. 

2°.  If  g  be  constant,  which  is  the  case  in  the  same  latitude, 
and  at  the  same  elevation  above  the  surface  of  the  earth, 

T  x  Vl; 
or,  the  time  will  vary  as  the  square  root  of  the  length. 

3°.  If  /  be  constant,  or  the  length  of  the  pendulum  remain 
the  same,  then 

T.-L 

vg 

or,  the  time  of  a  vibration  will  vary  inversely  as  the  square  rooi 
of  the  intensity  of  gravity. 


CONSTRAINED     MOTION.  185 

4°.  If  T  be  constant,  since 

lecg; 

or,  the  lengths  of  pendulums  vibrating  in  the  same  time  vary  as 
the  accelerating  force. 

304.  Cor.  Hence  the  force  of  gravity  in  different  latitudes, 
and  at  different  elevations  above  the  surface  of  the  earth,  will 
vary  as  the  length  of  the  pendulum  vibrating  seconds.  If, 
therefore,  the  length  of  the  seconds  pendulum  be  ascertained  at 
various  places  on  the  earth,  we  shall  have  the  relative  intensi- 
ties of  gravity  at  those  places.  Since  these  intensities  of  grav- 
ity at  different  places  are  dependent  upon  the  figure  of  the  earth, 
they  will  serve  to  determine  it.  The  pendulum,  therefore,  be- 
comes an  instrument  for  ascertaining  the  form  of  the  earth. 

305.  Prop.  The  lengths  of  pendulums  at  the  same  place  are 
inversely  as  the  square  of  the  number  of  vibrations  performed 
by  each  in  the  same  time. 

Let  N  and  N'  be  the  number  of  vibrations  performed  by 
each  respectively  in  the  time  H.  Then  the  duration  of  a  vibra- 
tion by  each  will  be 


T=|andT'=|, 

But 

T  :  T'=  Vl  :  V7=^  . 

H 

Hence 

1:  Z'=N'2  :  N2. 

Also, 

Z'N'2 
W' 

(87) 

306.  Sciiol.  To  determine  experimentally  the  length  of  the 
seconds  pendulum,  or  a  pendulum  which  will  oscillate  86400 
times  in  a  mean  solar  day,  let  a  clock,  the  length  /  of  whose 
pendulum  is  required,  be  regulated  to  vibrate  seconds.  Sus- 
pend in  front  of  the  clock  a  pendulum  of  known  length  /'(/'</), 
and  observe  the  exact  second  when  the  two  simultaneously 
commence  a  vibration.  As  the  vibration  of  /'  will  be  more 
rapid  than  that  of  I,  count  the  number  of  returns  to  coincidence 
of  vibration  in  a  period  of  five  or  six  hours ;  and,  finally,  note 


186  DYNAMICS 

the  exact  second  of  the  simultaneous  termination  of  a  vibra 
tion.  If  now  to  the  whole  number  of  seconds  N  shown  by  the 
clock,  there  be  added  twice  the  number  of  coincidences  of 
vibration,  we  shall  have  the  number  N'  of  vibrations  of  /'. 
These  values,  substituted  in  (87),  will  give  the  'ength  /  of  the 
seconds  pendulum. 

The  length  of  the  seconds  pendulum  in  the  latitude  of  New 
York  has  been  determined  to  be  equal  to 
39in.10168=3\25847. 

307.  Prop.  To  find  the  value  of  g,  the  measure  of  the  intensity 
of  gravity. 

From  (86),  we  deduce 

Making  T  =  l,  and  using  the  length  of  the  seconds  pendulum 

above,  we  find,  for  the  value  of  g  in  the  latitude  of  New  York, 

g-=385in.9183=32M598. 

308.  Prop.  To  find  the  correction  in  the  length  of  a  pendulum 
which  gains  or  loses  a  knovm  number  of  seconds  in  a  day. 

Let  /  be  the  length  of  the  seconds  pendulum,  w=86400, 
y~  the  number  of  seconds  gained  or  lost  in  a  day,  and  x=  the 
corresponding  correction  in  the  length. 

By  Art.  305,  and  recollecting  that  a  diminution  of  length 
corresponds  to  an  increase  in  the  number  of  vibrations, 

n2  :  (n+yy  —  l—x  :  I, 
from  which  we  deduce 

2ny+y'2 
X~n°+2ny+yt'  ' 
or,  rejecting  y2,  as  small  in  comparison  with  2ny, 

2y     . 

X  =  ~ ; ./ 

2y+n 

I 
=—  (88) 

l+2y 
Hence,  divide  the  length  of  the  seconds  pendulum  by  one  plus 


CONSTRAINED     MOTION.  187 

the  ratio  of  the  number  of  seconds  in  a  day  to  twice  the  gain  or 
loss,  and  the  quotient  will  be  the  correction  in  the  length. 

x  will  have  the  same  sign  as  y. 

309.  Prop.  To  determine  the  rate  of  clock  when  carried  to  a 
given  height  above  the  surface  of  the  earth. 

Let  N=86400,  N'=  the  number  of  vibrations  in  a  day  when 
the  clock  is  carried  to  the  height  h  above  the  surface  of  the 
earth,  and  r=  the  radius  of  the  earth.  The  length  of  the  pen- 
dulum remaining  the  same, 

T-T'=-L.-i-=I.l 

yf£  Jg>  N  N'' 

1        1 

°  r     (r+h) 

.:  N  :  N'=r+h  :  r, 

and  N' 


N-N'= 


r+h 

AN 


r+h 

N 
=— ;-  (89) 

1+h 

N— N',  the  loss  in  a  day,  is  the  rate  of  the  clock.  Hence, 
divide  the  number  of  seconds  in  a  day  by  one  plus  the  ratio  of 
the  radius  of  the  earth  to  the  height,  and  the  quotient  will  be  the 
rate.  The  quantities  r  and  h  must  be  in  the  same  denomina- 
tion. 

310.  Cor.  If  the  loss  in  a  day  by  a  seconds  pendulum  be 
determined  by  observation,  we  may  find  the  height  to  which 
it  is  carried,  for  (89)  gives 

.     r(N-N')  ,     x 

h=       N,       •  (90) 

311.  Def.  A  body  suspended  by  a  cord  and  performing 
revolutions  in  a  horizontal  circle  is  called  a  conical  pendulum. 

312.  Prop.   To  determine  the  motion  of  a  conical  pendulum. 
Let  /=  the  length  of  the  cord  AP,  fixed  at  A  and  attached 

to  the  body  at  P,  VC  =  r  the  radius  of  the  circle  which  the  body 


i88 


DYNAMICS. 


•describes  with  the  uniform  veloc- 
ity v,  m—  the  mass  of  the  body, 
and  the  angle  PAC  =  0.  Since 
the  body  is  in  the  same  circum- 
stances at  each  point  of  its  path, 
the  forces  acting  upon  it  must  be  in 
equilibrium.  These  are  the  ten- 
sion t  of  the  cord  in  the  direction 

Ail)2 
PA,  the  centrifugal  force /=—^-, 

in    the    direction    CP,    and    the 


weight  of  the  body  nig  downward. 
Resolving  horizontally  and  vertically,  we  have 

mv" 


t  sin.  6 — 


0. 


t  cos.O—mg  =0. 
From  (b),  we  have  the  tension  in  the  cord. 

mg 
cos.  d' 
Eliminating  t  from  (a)  and  (b), 

a    gr  sin.  d 


(a) 


=gl 


cos.  0 
sin.2  0 


cos.  d' 


The  time  t  of  performing  one  revolution  is 

_2nr 


cos.  6 


g 

AC 

g  ' 


313.  Dr.  Bowditch,  in  the  second  volume  of  his  translation 
of  the  Mecanique  Celeste,  gives  the  following  formula  for  com- 
puting the  length  /  of  the  seconds  pendulum  in  any  latitude  <f> 

/=A+wsin.20, 
in  which  A=39.01307  in.  and  w=0.20G44  in. 


CONSTRAINED     MOTION  189 

314.    EXAMPLES. 

Ex.  1.  The  length  of  an  inclined  plane  is  400  feet,  its  height 
250  feet :  a  body  falls  from  rest  from  the  top  of  the  plane. 
What  space  will  it  fall  through  in  3|  seconds?  in  what  time  will 
:t  fall  through  300  feet?  and  what  velocity  will  it  have  when  it 
arrives  within  50  feet  of  the  bottom  of  the  plane  ? 

Ex.  2.  The  angle  of  elevation  of  a  plane  is  25°. 30'.  A  body, 
in  falling  from  the  top  to  the  bottom,  acquires  a  velocity  of  450 
feet.     What  is  the  length  of  the  plane  ? 

Ex.  3.  The  length  of  a  plane  is  240  feet,  and  its  elevation  is 
36°.  Determine  that  portion  of  it,  equal  to  its  height,  which  a 
body,  in  falling  down  the  plane,  describes  in  the  same  time  it 
would  fall  freely  through  the  height. 

Ex.  4.  A  body  descending  vertically  draws  an  equal  body, 
25  feet  in  2|  seconds,  up  a  plane  inclined  30°  to  the  horizon 
by  means  of  a  string  passing  over  a  pulley  at  the  top  of  the 
plane.     Determine  the  force  of  gravity. 

Ex.  5.  The  time  of  descent  of  a  body  down  an  inclined 
plane  is  thrice  that  down  its  vertical  height.  What  is  the  in- 
clination of  the  plane  to  the  horizon? 

Ex.  6.  If  a  body  be  projected  down  a  plane  inclined  at  30° 
to  the  horizon  with  a  velocity  equal  to  fths  of  that  due  to  the 
vertical  height  of  the  plane,  compare  the  time  of  descent 
down  the  plane  with  that  of  falling  through  its  height. 

Ex.  7.  A  given  weight  P  draws  another  given  weight  W 
up  an  inclined  plane  of  known  height  and  length,  by  means  of 
a  string  parallel  to  the  plane.  When  and  where  must  P  cease 
to  act  that  W  may  just  reach  the  top  of  the  plane  ? 

Ex.  8.  Divide  a  given  inclined  plane  into  three  parts  such 
that  the  times  of  descent  down  them  successively  may  be 
equal. 

Ex.  9.  At  the  instant  that  a  body  begins  to  descend  down  a 
given  inclined  plane  from  the  top,  another  body  is  projected 
upward  from  the  bottom  of  the  plane  with  a  velocity  equal  tc 


190  DYNAMICS. 

that  acquired  in  falling  down  a  similar  inclined  plane  n  times 
its  length.     Where  will  they  meet  ? 

Ex.  10.  A  ball  having  descended  lo  the  lowest  point  of  a 
circle  through  an  arc  whose  chord  is  a,  impels  an  equal  ball 
up  an  arc  whose  chord  is  b.  Find  e  the  modulus  of  elasticity 
of  the  balls. 

Ex.  11.  Determine  that  point  in  the  hypothenuse  of  a  right- 
angled  triangle  whose  base  is  parallel  to  the  horizon,  from 
which  the  time  of  a  body's  descent  in  a  straight  line  to  the 
right  angle  may  be  the  least  possible. 

Ex.  12.  Two  bodies  fall  from  two  given  points  in  the  same 
vertical  line  down  two  straight  lines  to  any  point  of  a  curve  in 
the  same  time,  all  the  lines  being  in  the  same  vertical  plane. 
Find  the  equation  of  the  curve. 

Ex.  13.  193  oz.  is  so  distributed  at  the  extremities  of  a  cord 
passing  over  a  pulley,  that  the  more  loaded  end  will  descend 
through  3  inches  in  one  second.  What  is  the  weight  at  each 
end  of  the  cord  ? 

Ex.  14.  If  an  inelastic  body  be  constrained  to  move  on  the 
interior  of  a  regular  hexagon,  describing  the  first  side  with  a 
uniform  velocity  in  one  second,  find  the  time  of  describing  the 
last  side. 

Ex.  15.  A  stone  is  whirled  round  horizontally  by  a  string 
2  yards  long.  What  is  the  time  of  one  revolution,  when 
the  tension  of  the  string  is  4  times  the  weight  of  the  stone  ? 

Ex.  16.  What  is  the  length  of  a  pendulum  which  oscillates 
twice  in  one  second  ? 

Ex.  17.  A  seconds  pendulum,  carried  to  the  top  of  a  mount- 
ain, lost  48.6  seconds  in  a  day.  What  was  the  height  of  the 
mountain? 

Ex.  18.  The  length  of  a  pendulum  vibrating  sidereal  sec- 
onds, being  38.926  inches,  what  is  the  length  of  the  sidereal 
day  ?  How  much  must  it  be  lengthened  that  it  may  measure 
mean  solar  time  ? 


CHAPTER  VI. 


ROTATION     OF     RIGID     BODIES. 

315.  Prop.  When  a  rigid  body,  containing  a  fixed  axis,  is 
acted  upon  by  a  given  force  in  a  plane  perpendicular  to  that 
axis,  to  determine  its  motion. 

Let  the  annexed  figure  be  a  section  of  the 
body  by  a  plane  perpendicular  to  the  axis  at 
C,  and  the  given  force  F  act  at  B,  in  the  direc- 
tion AB,  at  a  given  distance  from  C.  If  v  be 
the  velocity  which  the  force  F  can  impart  to  a 
mass  M  when  free, 

F=Mu. 

When  the  force  acts  on  the  body,  each  parti- 
cle is  constrained  to  move  in  a  circle  whose  center  is  in  the 
axis  through  C.  If  w  be  the  angle  made  by  the  body  from  the 
action  of  the  force  in  a  unit  of  time,  the  linear  velocity  of  any 
particle  m  will  be  rw,  and  its  momentum  f=mro).  The  mo- 
ment of  this  force,  in  reference  to  the  axis  through  C,  is 
rf—mr^di.  The  moment  of  any  other  particle  will  have  the 
same  form,  and  hence 

rf+r'f'+,  &c,  =»ir2(,)+m'r'2w+,  &c, 
or  2.r/*=S.mr2w=w27?ir2, 

since  w  is  the  same  for  each  particle. 

But  the  moment  of  F  is  rF=Mu.CB,  and,  as  the  former  is  a 

measure  of  the  effect  of  the  latter, 

w2.mr=Mu.CB. 

Mu.CB 
.%  "=^ — r«  (°i) 

z.mr 
Thus,  if  the  body  consisted  of  two  particles  m  and  m'  at  the 
distances  r  and  r  from  the  axis,  and  the  force  F  would  impart 
to  m,  when  free,  the  velocity  v,  then  the  angular  velocity 


192 


D  Y  N  A  M  I  C  3. 


mvr 


mr  +m'r' 

and  the  linear  velocity  of  m  will  be 

mvifi 


mr%-\-m'rri' 
If  the  two  particles  were  equal,  and  at  equal  distances  from 
the  axis,  then  mr^-rn'r'2,  and 


Hm=m',  and  r'—2r, 


v 


v 

r«u=— . 

5 


Cor.  If  the  body  were  m  motion  before  the  force  acted  on 
it,  then  the  change  of  angular  velocity  will  be  given  by  (91). 

316.  Def.  The  moment  of  inertia  of  a  rigid  body  about  an 
axis  is  the  sum  of  the  products  of  the  mass  of  each  particle  by 
the  square  of  the  distance  of  that  particle  from  the  axis. 

Thus,  if  m  be  the  mass  of  a  particle  of  a  rigid  body,  r  its 
distance  from  a  fixed  straight  line,  2.??zr2  is  the  moment  of  in- 
ertia pf  the  body  about  that  line.  The  definition  given  above 
is  to  be  regarded  as  a  verbal  enunciation  of  this  analytical  ex- 
pression, which  has  required  nomenclature  by  the  frequency 
of  its'  occurrence  in  dynamical  investigations. 

317.  Prop.  The  moment  of  inertia  of  a  body  about  any  fixed 
axis  exceeds  its  moment  of  inertia  about  a  parallel  axis  passing 
the  center  of  gravity,  by  the  product  of  the  mass  into  the  square 
of  the  distance  between  the  axes. 

Let  m  be  a  particle  of  the  body,  and  let  the 
plane  ?nCG,  passing  through  m,  cut  the  two 
axes  at  C  and  G.  Through  the  axis  at  G 
pass  a  plane  perpendicular  to  CG,  intersect- 
ing the  plane  of  the  figure  in  AB.  Let 
mC=r,  mG=r„  and  CG=/*.  Draw  mF  per- 
pendicular and  niD  parallel  to  AB.  Then 
r=r;+/(2±2/i.GD, 
according  as  the  angle  mGC  shall  be  obtuse  or  acute.     Hence 


ROTATION     OF     RIGID     BODIES. 


193 


mr^=mr\+mlii±2mh.GT>, 
and  2.mr*=2.mr21+h22m±2h2.m.GT). 

Now,  since  the  plane  through  the  axis  at  G  passes  through 
the  center  of  gravity,  by  (29), 

2.m.GD=2.m.Fm=0. 

.-.  2.mr2=2.mr?+A22m 
=Zmr\+Mh\ 
M  being  the  whole  mass  of  the  body. 

318.  Schol.  Assuming  k,  such  that  Mki=2.mra1,  we  have 

2.mr2=M(F+A2). 
MA-2  is  the  moment  of  inertia  of  a  body  about  an  axis  through 
the  center  of  gravity,  and  M(F+/j2)  is  the  moment  of  inertia 
about  a  parallel  axis  at  a  distance  h  from  the  former.  The 
moment  of  inertia  about  any  axis  will,  therefore,  be  easily  de- 
termined when  the  moment  of  inertia  about  an  axis  through 

i  r  ■      •    i  «■  2.mr2    . 

the  center  of  gravity  is  known.  Since  k  —  M  ,  the  determ- 
ination of  k  will,  in  general,  require  the  aid  of  the  integral  cal- 
culus. « 


The  length  VA2+/j2  is  called  the  radius  of  gyration  about 
the  axis  considered,  and,  similarly,  k  is  called  the  radius  of 
gyration  about  an  axis  through  the  center  of  gravity.  Since 
k  is  the  least  value  of  V/f+h*,  it  is  sometimes  called  the  prin- 
cipal radius  of  gyration. 

319.  Prop.  If  a  body  oscillate  about  a  fixed  horizontal  axis 
not  passing  through  its  center  of  gravity,  there  is  a  point  in  the 
right  line,  drawn  from  the  center  of  gravity  perpendicular  to  the 
axis,  whose  motion  is  the  same  as  it  would  be  if  the  whole  mass 
were  collected  at  that  point  and  allowed  to  vibrate  as  a  pendulum 
about  the  fixed  axis. 

Let  the  horizontal  axis  be  perpendicular  to 
the  plane  of  the  figure  at  C,  G  be  the  center 
of  gravity,  CA  be  horizontal,  and  GA  vertical. 
Then,  if  M  be  the  mass  of  the  body,  and  v  the 
velocity  which  gravity  can  impart  to  the  body 
if  free  in  each  instant  of  time,  the  moment  of 

N 


194  DYNAMICS. 

gravity  will  be  Mu.CA.     By  Art.  315,  Cor.,  tne  change  of  an- 
gular velocity  produced  in  each  instant  of  time  is 

_Mi?.CA 
2.mra  ' 
Produce  CG  to  some  point  O  and  draw  OB  vertically,  meeting 
CA  in  B.  Now,  if  the  whole  mass  of  the  body  were  collected 
at  0,  the  moment  of  inertia  of  the  body  would  be  M.CO3,  and 
the  change  of  angular  velocity  in  the  same  instant  of  time 
would  be 

w'~"M.coa' 

But  the  position  of  the  point  O  being  arbitrary,  CO,  and  there- 
fore CB,  will  vary  at  pleasure.     We  may,  therefore,  put 
Z.mr'  :  M.C03=CA  :  CB=CG  :  CO,  (a) 

which  gives 

CA         CB 


S.mr*     M.CO" 
or  w=w' ; 

that  is,  the  change  of  angular  velocity  is  the  same  in  the  two 
cases,  and,  therefore,  the  motion  from  rest  will  be  the  same. 
320.  Cor.  The  point  O  may  be  found  from  (a),  which  gives 
S.mr*     M(&2+/i2) 


CO: 


M.CG-      M.h 

_k2+h2 
h 


=h+j.  (92) 

321.  Def.  The  point  O  is  called  the  center  of  oscillation  of 
the  body  with  respect  to  the  axis  through  C.  It  is  thus  de- 
fined :  when  a  rigid  body  moves  about  a  fixed  horizontal  axis 
under  the  action  of  gravity,  in  the  straight  line  drawn  through 
the  center  of  gravity  perpendicular  to  the  axis,  a  point  can  be 
found  such  that,  if  the  mass  of  the  body  were  collected  there 
and  hung  by  a  thread  from  the  axis,  the  angular  motion  of  the 
point  would,  under  the  same  initial  circumstances,  be  the  same 
with  that  of  the  body,  and  this  point  is  called  the  center  of 
oscillation  of  the  body  with  respect  to  the  axis. 


ROTATION     OF     RIGID     BODIES.  195 

322.  Cor.    Her.ce,  when  a  body  makes  small  oscillations 
about  a  fixed  horizontal  axis,  it  is  only  necessary  to  calculate 

the  value  of  the  expression  /j+-t-=/,  and  the  time  of  a  small 


oscillation  will  be 


*\J1--     (86). 


323.  Def.  A  body  of  any  form  suspended  from  a  fixed  axis, 
about  which  it  oscillates  by  the  force  of  gravity,  is  called  a 
compound  pendulum. 

324.  Prop.  The  centers  of  oscillation  and  suspension  are  con- 
vertible. 

Let  h'  be  the  distance  of  the  center  of  oscillation  from  the 
center  of  gravity,  when  h  is  the  distance  of  the  axis  from  the 
center  of  gravity.     Then,  by  (92), 

GO-h=h'=j, 

and  h=- ;  (93) 

so  that  if  h'  be  the  distance  from  the  center  of  gravity  to  the 
axis,  h  will  be  the  distance  from  the  center  of  gravity  to  the 
center  of  oscillation,  and  a  body  will  oscillate  in  the  same  time 
about  an  axis,  through  the  center  of  oscillation,  as  it  oscillates 
about  the  original  axis,  the  extent  of  vibration  being  the  same. 

325.  Cor.  1.  If  in  a  straight  line  through  the  center  of  gravity, 
perpendicular  to  the  axis  of  motion,  points  be  taken  at  distances 
h  and  h'  from  the  center  of  gravity,  and  on  opposite  sides  of  it, 
then  the  length  of  the  equivalent  simple  pendulum  is  h+h' ;  so 
that  the  time  of  vibration  about  the  axes  through  each  of  these 
two  points  is  the  same,  and  the  length  of  the  equivalent  simple 
pendulum  is  in  each  case  the  distance  between  the  two  axes. 
Each  of  these  points  is  the  center  of  oscillation  in  reference  to 
the  other  as  a  center  of  suspension. 

326.  Cor.  2.  From  (93),  we  have 

hh'=1?\ 

or,  the  principal  radius  of  gyration  is  a  mean  proportional  /»; 


196  DYNAMICS. 

tween  the  distances  of  the  centers  of  oscillation  and  suspension 
from  the  center  of  gravity. 


327.  Cor.  3.  Since  h  :  Vk*+h*=  Vk*+h*  :h+-r=CO,  we  in* 

h 

fer  that  the  distances  of  the  centers  of  gravity,  of  gyration,  and  of 
oscillation  from  the  axis  of  motion,  are  continued  proportionals. 

328.  Schol.  The  convertibility  of  the  centers  of  oscillation 
and  suspension  was  employed  by  Captain  Kater  in  finding  the 
length  of  a  simple  pendulum  vibrating  seconds,  and  consequent- 
ly the  force  of  gravity  at  the  place  of  observation. 

A  bar  of  brass  one  inch  and  a  half  wide  and  one  eighth  of  an 
inch  thick  was  pierced  by  two  holes,  through  which  triangular 
wedges  of  steel,  called  knife  edges,  were  inserted,  so  that  the 
pendulum  could  vibrate  on  the  edge  of  either  of  these  as  an  axis, 
resting  on  two  fixed  horizontal  plates  of  agate,  between  which 
the  bar  was  suspended.  The  axes  were  about  39  inches  apart. 
Weights  were  attached  to  the  bar  and  rendered  capable  of 
small  motions  by  screws,  by  which  means  the  position  of  the 
center  of  gravity  of  the  bar  could  be  changed.  These  weights 
were  so  adjusted  by  trial  that  the  time  of  a  small  vibration 
through  an  angle  of  about  1°  was  the  same  when  either  knife 
edge  was  the  axis,  so  that  each  gave  the  center  of  oscillation 
belonging  to  the  other.  The  distance  of  the  knife  edges  was  ob- 
tained by  placing  the  pendulum  so  that  the  edges  were  viewed 
by  two  fixed  microscopes,  each  furnished  with  a  micrometer  of 
ascertained  value,  and  afterward  placing  a  scale  of  known  ac- 
curacy in  a  similar  position  under  the  same  microscopes.  This 
method,  combined  with  that  referred  to  in  Art.  306,  served  to 
determine  the  length  of  the  seconds  pendulum,  and  thence  the 
force  of  gravity. 

329.  The  relation  of  the  simple  to  the  compound  pendulum 
will  be  illustrated  by  one  or  two  examples  in  which  the  prin- 
cipal radius  of  gyration  is  supposed  to  have  been  previously 
determined. 

Ex.  1.  A  material  straight  line  vibrates  about  an  axis  per- 
pendicular to  its  length :  required  the  length  of  the  isochronal 
simple  pendulum. 


ROTATION     OF     RIGID     BODIES.  197 

Let  2a  be  the  length  of  the  line,  and  h  the  distance  of  the 
point  of  suspension  from  its  center  of  gravity,  which  is  its  mid- 
dle point.  The  radius  of  gyration  of  a  straight  line  about  an 
axis  through  its  center  of  gravity,  perpendicular  to  its  length, 

a 
is  ~7r.     Since  the  radius  of  gyration  is  a  mean  proportional 

between  the  distances  of  the  centers  of  oscillation  and  of  sus- 
pension from  the  center  of  gravity  (93), 

vJL* 

h    3k' 

2 

and  {Art.  325),  l=h+h'=h+-r. 

Oil 

1°.  Let  the  point  of  suspension  be  at  the  extremity  of  the 
line,  in  which  case  h=a. 

Then  h'=\a, 

and  /=a  +  i«  =  |.2«, 

that  is,  the  center  of  oscillation  is  two  thirds  of  its  length  below 
the  axis  of  motion. 

2°    Let  h=\a;  then 

h'=\a 
and  Z=|.2a, 

the  same  as  before.  Hence  the  time  of  a  small  vibration  is 
the  same,  whether  the  line  be  suspended  from  one  extremity, 
or  from  a  point  one  third  of  its  length  from  the  extremity. 
This  also  illustrates  the  convertibility  of  the  centers  of  oscilla- 
tion and  of  suspension. 

3°.  If  h=%a, 

h'=$a, 
and  Z=}fa=f.2a+i.2a; 

or,  when  the  center  of  suspension  is  three  eighths  of  its  lengtn 
from  one  end,  the  center  of  oscillation  is  one  sixth  of  its  length 
below  the  other  end. 

4°.  lfh=0,  h'=ao, 

and  1=  ao  ; 

or,  when  the  center  of  suspension  is  at  the  center  of  gravity, 
the  length  of  the  equivalent  simple  pendulum  is  infinite,  and 


198  DYNAMICS. 

therefore  the  time  of  one  vibration  is  infinite.  This  is  obvious, 
also,  from  the  fact  that,  when  the  line  is  suspended  from  the 
center  of  gravity,  no  motion  can  result  from  the  action  of 
gravity. 

Ex.  2.  A  sphere  being  made  to  oscillate  about  a  given  axis  : 
required  the  length  of  the  equivalent  simple  pendulum. 

Let  r  be  the  radius  of  the  sphere,  and  h  the  distance  of  the 
axis  of  motion  from  its  center.  The  principal  radius  of  gyra- 
tion of  a  sphere  is  rV%. 

2r2 
Hence  h'——r . 

bh 

1°.  Let  h=r,  or  the  spnere  vibrate  about  an  axis  tangent  to 
its  surface.     Then 

&'=§r, 

and  /=r+|r; 

or,  the  center  of  oscillation  is  two  fifths  of  the  radius  distant 
from  the  center  of  the  sphere. 

r 


2°.  If/*=10r, 

4'=35' 

and 

;=10''+i 

3°.  If  h=\r, 

h'=2r, 

and 

r 
o 

or,  the  center  of  oscillation  is  without  the  sphere  and  at  a  dis- 
tance r  from  the  surface. 


MOMENT   OF   INERTIA. 

330.  Since  {Art.  316)  the  expression  for  the  moment  of  in 
ertia  of  a  system  consisting  of  a  finite  number  of  points  is  2.mr*, 
m  being  the  portion  of  the  mass  which  is  at  the  distance  r  from 
the  axis  of  rotation ;  when  the  number  of  points  becomes  in- 
definite, the  expression  will  evidently  become  fr'dM,  dM  be- 
ing an  element  of  the  mass  at  the  distance  r. 

Hence  (Art.  318),        Mk*=fr*dM.  (a) 


ROTATION     OF     RIGID     BODIES.  199 

We  shall  now  illustrate  the  method  of  determining  M/ta  in  a 
few  simple  cases. 

Ex.  1.  To  find  the  moment  of  inertia  of  a  straight  line  re 
volving  about  an  axis  perpendicular  to  it  at  any  point  of  it. 

Let  the  axis  be  at  a  distance  a  from  one  end  and  b  from  the 
other/  and  let  r  be  any  distance  from  the  axis ;  since  the  thick- 
ness and  density  of  the  line  are  supposed  to  be  uniform,  eacl 
may  Vvj  taken  =1. 

.-.  M=a+b  and  dM=dr. 

r3 
Hence  (a)  (a+&)&2==/VWr=— +c, 

o 

and  the  integral  being  taken  between  the  limits  —a  and  +b. 

.      v=  aa+b° 

3(ci+b)' 
If  a=b,  or  the  axis  be  at  the  center  of  gravity,  we  have,  for 
the  principal  radius  of  gyration, 

a 

h=V3' 

__„     Ma2    2  , 

and  MF=— =-a3, 

o        o 

whatever  be  the  thickness  and  density  of  the  line. 

Ex.  2.  To  find  the  moment  of  inertia  of  the  circumference 
of  a  circle  about  an  axis  perpendicular  to  its  plane  through  its 
center. 

In  this  case  M=27rr,  and  since  all  the  points  are  at  the  same 
distance  from  the  axis,  r  is  constant. 

...  Mk^fridM=rydM=2nr3, 
and  k=r. 

Ex.  3.  To  find  the  moment  of  inertia  of  a  circle  about  an 
axis  through  the  center  perpendicular  to  its  plane. 

Putting  a=  the  radius  of  the  circle,  its  area  will  be  na?,  and 
at  a  distance  r  from  the  axis  the  area  will  be  7rr2.  Hence,  in 
this  case,  dM=2nrdr  and  MAa=7raT=2rr/r*rfr=$7rr4=£7ra4, 
when  r=a,  and 


200  DYNAMICS. 

Ex.  4.  To  find  the  moment  of  inertia  of  the  circumference 
of  a  circle  about  a  diameter. 

Taking  the  proposed  diameter  for  the  axis  of  a:,  the  distance 
of  any  point  (x,  y)  from  the  axis  of  rotation  is  y. 

From  the  equation  of  the  circle,  yi=a2—x',  we  find 


Vdxi+dyi=dM=-dx. 

y 

.:  r-dM=y*dM=aydx, 
and  Mk'i=2nak'i=afydx=aX  area  of  circle=Tra9. 

a 

' "     ~V2' 

Ex.  5.  To  find  the  moment  of  inertia  of  a  sphere  about  a 
diameter. 

Let  the  axis  of  rotation  be  the  axis  of  x,  and  conceive  the 
sphere  to  be  generated  by  a  circle  of  variable  radius,  y,  whose 
center  moves  along  the  axis.  By  Ex.  3  the  moment  of  this 
generating  circle  is  ^y\  Hence  the  whole  moment  will  be 
the  sum  of  the  moments  of  the  generating  circle  in  all  its  posi 
tions. 

.-.  Mk'i=\7ifyidx. 

Buty^tf-x*.         .'.  Mki=\-nj\ai-xiydx, 
and  this  integral,  taken  between  the  limits  +a  and  —a,  is 

But  M=A*a'.  .-.  A=aV§. 


CHAPTER  VII. 

331.  The  methods  employed  in  the  preceding  chapters  to 
determine  the  motion  of  a  body  are  partial  in  their  application 
and  limited  to  the  simplest  cases.  When  the  force  is  variable, 
or  the  motion  of  a  body  is  due  to  the  action  of  several  forces, 
varying,  in  direction  as  well  as  in  intensity,  with  the  varying 
position  of  the  body,  some  more  general  method  is  required. 
We  now  proceed  to  show,  to  a  limited  extent,  how  the  circum- 
stances of  the  motion  of  a  body  or  point,  in  such  cases,  may  be 
determined.  In  order  to  this  the  fundamental  formulas  will 
need  some  modification  and  extension. 

332.  By  Art.  214,  variable  velocity  at  any  instant  is  meas- 
ured by  the  limit  of  the  ratio  of  the  space  to  the  time,  that  is, 
from  the  nature  of  the  differential  coefficient,  it  is  the  differen- 
tial coefficient  of  the  space  regarded  as  a  function  of  the  time 
Hence 

In  like  manner,  it  appears  from  Art.  223,  that  the  force,  when 
variable,  is  measured  at  any  instant  by  the  differential  coeffi- 
cient of  the  velocity  regarded  as  a  function  of  the  time.    Hence 

*=-.  [VIII.] 

From  [VII.],  by  differentiation,  we  obtain 
dv    <fs 
Tt~W 

•••  *-y  [IX-l 

From  [VII.]  and  [VIII.]  we  obtain,  "by  eliminating  dt, 
vdv~<pds.  [X.j 


202  DYNAMICS. 

§  I.    RECTILINEAR    MOTION    OF    A    FREE    POINT. 

333.  If  a  free  point  at  rest  is  acted  on  by  a  single  force,  or 
forces  whose  resultant  is  equivalent  to  a  single  force,  the  mo- 
tion must  be  entirely  in  the  direction  of  that  force,  and  may  at 
once  be  determined  from  [VII.]  and  [VIII.].  But  in  order 
that  we  may  integrate  these  expressions,  the  law  of  variation 
of  the  force  must  be  given  as  a  function,  either  of  the  space, 
time,  or  velocity ;  and  as  the  intensity  of  a  variable  force 
would  depend  on  the  position  of  the  body,  the  force  is  natu- 
rally and  generally  given  as  a  function  of  the  distance  of  the 
body  from  some  fixed  point. 

In  order  that  we  may  obtain  an  exact  expression  for  the 
value  of  the  force  at  any  variable  distance,  its  value  at  some 
given  distance  must  be  known.  This  given  distance  is  most 
conveniently  assumed  as  the  unit  of  distance,  and  if  p  be  the 
intensity  of  the  force  at  this  unit  of  distance,  it  is  usually  called 
the  absolute  force. 

If,  then,  the  force  be  given  to  vary  as  the  n"'  power  of  the 
distance  z  from  the  fixed  point,  we  have 

<(>=[ixn.  (94) 

We  shall  first  consider  the  case  of  a  constant  force,  for  the 
purpose  of  showing  with  what  facility  the  calculus  enables  us  to 
determine  the  relations  of  the  time,  space,  and  velocity,  although 
these  relations  have  already  been  deduced  in  chapter  III. 

334.  Prop.  To  determine  the  space  described  by  a  point  acted 
on  by  a  constant  force  in  terms  of  the  time,  the  velocity  of  the 
point  in  the  direction  of  the  force  at  the  commencement  of  the 
time  being  given. 

The  velocity  due  to  the  action  of  the  force  in  the  time  t  is, 
by  [III.], 

where  </>  represents  the  velocity  generated  by  the  force  in  the 
unit  of  time. 

If  vl  be  the  given  velocity,  or  the  velocity  when  the  time 
commences,  we  have,  for  the  whole  velocity,  as  in  (59), 

v=v1±(}>t. 


RECTILINEAR     MOTION     OF     A     FREE     POINT.     203 

Substituting  this  value  of  v  in  [VII.],  we  have 

ds=v  ldt±(ptdt, 
and,  integrating, 

in  which  c  is  an  arbitrary  constant  depending  on  the  position 
of  the  point  when  2=0. 

If  the  space  be  reckoned  from  the  position  of  the  point  when 
t=0,  then  c=0,  and 

s=v1t±l(pf.     (60). 
If  the  point  move  from  rest,  vl=0,  and 
s=%>t\     (54).      - 
All  the  other  relations  are  readily  deduced  from  these. 

335.  Prop.  To  determine  generally  the  velocity  of  a  particle 
?noving  in  a  right  line  to  or  from  a  fixed  point  by  the  action  of 
a  variable  force.  • 

By  [X.],  vdv=(f>ds ; 

or,  measuring  the  line  on  the  axis  of  x, 

vdv=<pdx. 
Hence,  by  integration, 

v<2=2f(pdx+c;  (95) 

and  if  the  force  varies  as  some  function  of  the  distance  x,  the 
integration  may  in  general  be  effected. 

If  the  motion  is  toward  the  fixed  point,  dx  will  be  negative ; 
if  from  it,  positive. 

To  determine  the  constant,  we  must  know  the  velocity  at 
some  given  point. 

336.  Prop.   To  determine  generally  the  time  of  the  motion. 

From  [VII.]  we  have  at  once 

~dx  •  .     . 

t=f— +c,  (96) 

*    v 

and  since  v  is  determined  by  (95)  in  terms  of  x,  the  integra- 
tion may  in  general  be  effected. 

To  determine  c,  we  must  know  the  position  of  the  particle 
at  some  given  time. 


204  DYNAMICS. 

337.  Prop.  To  determine  the  velocity  of  a  particle  attracted 
to  a  fixed  point  by  a  force  varying  directly  as  the  distance  of  the 
particle  from  that  point. 

Assuming  the  axis  of  x  in  the  direction  of  the  motion,  anf* 
the  origin  at  the  center  of  force,  by  (94)  we  have 

<f>=  —  \ix; 
the  sign  being  negative,  because  the  force  which  is  directed  to 
the  origin  tends  to  diminish  the  distance  x. 

Substituting  this  value  of  </>  in  (95),  and  integrating,  we  get 

vi=c— [IX1. 

If  a  be  the  distance  of  the  particle  from  the  origin  when  the 
motion  commences,  when  x=a,  v=0.     Hence 

and  v'=fi(aa-tt9),  97) 

from  wjiich  it  appears  that  x  can  not  pass  the  limits  ±a. 

338.  Cor.  Since  the  force  by  which  a  material  point,  at 
iiberty  to  move  along  a  perforation  from  the  surface  to  the 
center  of  the  earth,  varies  directly  as  the  distance  from  the 
center,  (97)  will  serve  to  determine  the  velocity  of  the  particle 
at  any  distance  x  from  the  center. 

In  this  case,  if  a=r,  the  radius  of  the  earth,  and  g  the  force 
of  gravity  at  the  surface,  \i  will  be  found  from  the  proportion 
r:  l=g:p. 

Hence  u2=^(r5-.<). 

r 

As  this  velocity  must  be  spent  before  the  particle  stops,  if  we 
make  v=0,  we  get  x=r,  or  the  particle  will  go  to  the  opposite 
point  of  the  earth's  surface. 

339.  Prop.  To  find  the  time  when  the  force  varies  directly  as 
vfie  distance. 


dx 
From  [VII.],  we  have      dl= — 


dx 


V[J>  Va?—x*' 
To  integrate  this,  let  x=a  sin.  6,  whence  a1— xt=a'i  cos.'B. 


RECTILINEAR     MOTION     OF     A     FREE     POINT.      205 

1     a.d  sin.  6 


\  dt=- 


i/fi   a  cos.  u 

=-L  de, 

and  t=—=.6+c. 

Hence  6=  V[t(t—c), 

and  x=a.s'm.V[t(t—c). 

Suppose  the  time  to  commence  when  the  pai'icle  is  at  tne 
origin,  or  x=0;  then 

0=—  a.sin.  Vp.c, 
or  c—0. 

.-.  t=^.d.  {a)' 

lfx=a,  sin.  0=1,  and  d=±n  or  §tt,  &c, 

1  1      o 

and  t=±n—=  or  #7r-^=,  &c, 

V[j>       "    Vf* 
from  the  commencement  of  the  motion. 

If  x=  —  a,  sin.  0=  —  1,  and  0=§7r  or  |7r,  &c, 

1  1      . 

and  i=^7r — =  or  ^tt^=,  &c. 

Hence  the  particle  moves  from  +a  to  —  a,  while  t  changes 

from  |7r— r=  to  f  tf-7=f  or  the  time  of  one  vibration  is 
Vp       "    vjti 

T=tt-4=,  (98) 

and  this  is  true  whether  a  be  large  or  small.     Compare  Art. 
302. 

Cor.  1.  When  a  particle  is  attracted  to  a  fixed  point  by  a 
force  varying  directly  as  the  distance,  the  time  of  descent  to 
that  point  will  be  the  same  for  all  distances. 

Cor.  2.  The  time  of  descent  to  the  center  of  the  earth,  in 

g 
which  case  w==-.  is 
r 


200 


DYNAMICS. 


m      it     /r 

T'=2\/v 


Q 


=2i'.5".a 

Cor.  3.  Let  S  be  the  center  of  force,  AS=a 
the  distance  of  the  particle  at  the  commence- 
ment of  motion,  P  the  position  of  the  particle  at 
any  instant,  and  t  the  time  of  describing  PS=a;. 
On  AS  describe  a  quadrant,  and  draw  PQ  per- 
pendicular to  AS.     Then 

£=a.sin.QSB. 
.-.  QSB=0  and  QB=«.0. 

V>.  V~rf^a?=  V> .PQ  a  PQ, 
1    „      1    QB_     QB       _nn 

a 


By  (97), 

and  by  (a), 


t=- 


V[i         Vfi'   a       vel- atS 
When  the  particle  is  at  A,  arc  BQ  becomes  BA 


and 


i :rrv  = 

2v> 


340.  Prop.   To  determine  the  velocity  when  a  particle  is  at- 
tracted by  a  force  varying  inversely  as  the  square  of  the  distance. 


In  this  case  (94), 
and  [X], 
Integrating, 
If  v=0  when  x=a, 


<p=     fix    , 
vdv=—[ix~~dx. 
r2=     2/«T1+c. 
c=—  2fia~\ 

\  v*=2n(x-1-a-1)=2[i 


a—x 


(99) 


341.  Cor.  Since  the  intensity  of  gravity  above  the  earth's 
surface  varies  directly  as  the  mass  of  the  body,  and  inverse- 
ly as  the  square  of  its  distance  from  the  center,  (99)  will 
give  the  velocity  of  a  body  falling  from  any  height  to  the 
surface.     When  the  body  arrives  at  the  surface  x=rf  and 

1      1 

j,  t*=gr 


since  ju  :  g=- 


V     r 


Hence 


v"=2gr.- 


If  a=cc 


v—  V2gr=6.9428  miles, 


HECTILINEAR     MOTION     OF     A     FREE     1UINT.      207 

*'rom  which  it  appears  that  the  velocity  can  never  equal  seven 
miles,  and  if  a  body  be  projected  upward  with  the  above  ve- 
locity, supposing  no  resistance  from  the  air,  it  would  never  re- 
turn. 

If  we  make  x=0  in  (99), 

u=ao  ; 
or  the  velocity  at  the  center  would  be  infinite  if  the  same  law 
of  force  continued. 

342.  Prop.   To  find  the  time  when  the  force  varies  inversely 
as  the  square  of  the  distance. 

We  have,  from  [VII.], 

—  dx           —dx 
dt= = 

/  a  \  2  /*  —xdx 
"  '     \2J*/t/  Vax-x*' 
By  adding  \adx  to  the  numerator,  and  then  subtracting  the 
same  quantity  from  it,  we  have 

—  (  a  \* PS  \adx—xdx        \adx      1 
\2fxJ  J    (    Vax—x2       Vax—x2  > 

~l^r-      \{ax— x  r  — -versin. \-c\. 

\2fi/    iV  '      2  a        s 

_2x 
When  t=0,  x=a,  and  versin.  J — =tt. 

a 

(a  \ '  la  2x\  ) 

-)   {(ax-xy+-(ir-verBm.-*—)\.   (100) 

Cor.  1.  If  x=0,  we  have 

or  the  square  of  the  time  of  falling  to  the  center  varies  as  the 
cube  of  the  distance. 

Cor.  2.  On  AS=a  describe  a  semicircle,  and  draw  PQ,  per- 
pendicular to  AS.     Then  PQ=  Vax--x*,  arc  AQ,S=47m,  arc 

SQ=SC.z.SCQ=|  versin.-^. 


208  DYNAMICS. 

Hence  the  time  through  AP  is,  by  (100), 
x 

t=  (J-\  2(PQ+  arc  AQS-  arcQS) 
i 

|^(PQ+arcAQ). 

If  SQ,  be  produced  to  meet  in  R  a  tangent  at  A* 
SA.PQ    aVax—x2 


AR= 


PS 

'2p 


/a—x\ 
\  ax  ) 


.  by  (99),       v=  f-Pj  2 .  AR  ac  AR. 

When  P  coincides  with  S,  AR  is  infinite. 

343.  Prop.  To  find  the  velocity  and  time  when  the  force  varts* 
inversely  as  the  cube  of  the  distance. 


In  this  case, 

<j)=      (IX   3, 

and  [X], 

vdv=—[ix~3dx. 

Integrating, 

v2=     [ix~*-\-c. 

But  when  x=a,  v—0  ; 

.'.    c——iia~i, 

and 

u2=     [i(x~*— a-2), 

or 

,-  Va'-x2 

v=  vu. . 

ax 

By  [VII.], 

dx      a      — xdx 

dt —         — " 

v        Vf*  Va^—x* 
a 

Integrating, 

t=-^.Va2-x*+c. 

When  x=a,  t=0 ; 

.:  c=0, 
a 

and 

t=—=.Va2-x\ 

V/i 

a' 

lfar=0, 

VfJ 

(101) 


(102) 


the  time  to  the  center  of  force. 

«      x,.                   v  .«     AS.PQ    aV7^? 
Ccr.  (See  Fig.,  Art.  339)  AR=     PS    = • 

V* u, 

.-.  v=— =-.  AR  ac  AR, 

a 

and  t=~=  PQocPQ. 

V/i 


CURVILINEAR     MOTION     OP     A     FREE     POINT.     209 


§   II.    CURVILINEAR   MOTION   OF   A   FREE   POINT. 

344.  If  several  forces  in  different  directions  in  the  same 
plane  act  continually  on  a  material  point,  it  will  have  a  result- 
ing motion  at  any  instant,  whose  direction  and  rate  must  be  de- 
termined by  the  relations  between  the  directions  and  intensi- 
ties of  the  forces  which  act  upon  it.  The  motion  may  or  may 
not  be  rectilinear,  and,  in  order  to  investigate  the  circumstances 
of  the  resulting  motion  generally,  we  shall  employ  the  method 
of  resolution  to  two  rectangular  axes,  as  in  Statics,  using  X 
and  Y,  instead  of  2.X  and  2.Y,  to  denote  the  sums  of  the  re- 
solved forces. 

345.  Prop.  To  determine,  generally,  the  motion  of  a  point  act- 
ed upon  by  any  number  of  forces  in  the  same  plane. 

Resolve  the  forces  in  the  direction  of  two  rectangular  axes, 
and  let  X  and  Y  represent  the  sums  of  the  resolved  forces  in 
each  axis,  x  and  y  being  the  co-ordinates  of  the  position  of  the 
point.  The  point  may  be  regarded  as  acted  upon  by  the  two 
forces  X  and  Y,  independently  of  each  other.    Hence,  by  [IX.], 

__     d*x    ^T    d2y 

If  ds  represent  any  small  element  of  the  path  of  the  point,  by 

ds 
[VII.],  -j-  will  represent  the  velocity  of  the  point  in  its  path. 

Hence,  if  a  and  /3  represent  the  angles  which  ds  makes  with 
the  axes,  the  velocities  in  the  direction  of  the  axes  (39)  are 

ds  dx   ds  n    dy 

Tfcos-  a=di<  dfcos-  p=di-         <104> 

Such  are  the  general  equations  by  which  the  motion  c  f  a 
point  in  a  plane  may  be  determined. 

346.  If  the  velocity  is  given,  we  must,  by  resolution,  obtain 
the  velocity  in  each  axis,  and  thence,  by  differentiation,  the 
force  in  each  axis,  and,  by  composition,  the  resultant  force 
may  be  determined. 

If  the  forces  are  given,  we  may,  by  integration,  find  the  ve- 

O 


210  DYNAMICS. 

locity  in  each  axis,  and  thence,  by  composition,  the  velocity  of 
the  point  in  its  path. 

If  the  path  is  required,  each  of  equations  (103)  must  under- 
go two  integrations,  at  each  of  which  operations  a  constant 
must  be  introduced.  The  constants  introduced  at  the  first  in- 
tegration will  depend  on  the  velocity  of  the  point  when  the 
time  commences,  or  at  some  given  time ;  those  introduced  at 
the  second  integration  will  depend  on  the  position  of  the  point. 
We  thus  have  two  equations  involving  the  co-ordinates  x  and 
y  and  the  time  t,  and,  by  elimination  off,  an  equation  between 
the  co-ordinates  will  be  obtained,  which  is  the  equation  of  the 
path  of  the  point. 

347.  Prop.   To  determine  the  velocity  of  a  point  in  its  path. 
The  velocity  may  be  found  as  above  indicated,  but  the  fol- 
lowing is  the  preferable  method. 

Multiply  the  first  of  equations  (103)  by  2dx,  the  second  by 
2dy,  and,  adding  the  results,  we  have 

The  first  member  of  this  equation  is  the  differential  of 
dxiJrdy"     ds2       a 

~~de    =de=v' 

Therefore,  integrating,  we  find 

v*=2f(Xdx+Ydy)+c.  (105) 

348.  Cor.  If  this  expression  ~K.dx-\-Ydy  is  integrable,  the  ve- 
locity may  be  found,  provided  we  can  correct  the  integral,  or 
know  the  velocity  at  some  given  point. 

Thus,  if  the  expression  is  a  differential  of  the  co-ordinates 
of  the  position  of  the  point,  so  that  its  integral  is  a  function  of 
these  co-ordinates,  or 

«9=2/(*,y)+c, 
and  if  u,  is  the  velocity  at  the  point  whose  co-ordinates  are  a 
and  b,  so  that 

v\=2f{a,b)+c; 
then  v* -  v\ = 2f(x,  y)  -  2f(a,  b) . 

Hence  it  appears  that  the  velocity  acquired  by  the  particle 


CURVILINEAR     MOTION     OF     A     FREE     POINT.      211 

in  passing  from  one  point  (a,  b)  to  any  other  (x,  y)  is  the 
same,  whatever  be  the  curve  described  between  these  points, 
since  the  change  of  velocity  is  independent  of  the  co-ordinates 
of  any  intermediate  point. 

349.  Prop.  The  expression  Xdx+Ydy  is  always  integrable 
whenever  the  forces  are  directed  to  fixed  centers,  and  their  intens- 
ity is  a  function  of  the  distance  from  those  centers. 

Let  F,  Fj  .  .  .  be  two  forces  directed  to  fixed  centers, 
"    a,  b,  al,  &j  be  the  co-ordinates  of  the  centers, 
"    x,  y  ....  be  the  co-ordinates  of  any  position  of  the 

particle, 
"         r,  rx    .  .  be  the  distances  of  the  particle  from  the  cen- 
ters, 
"    a,(3,a1,(51  be  the  angles  which  r  and  r,  make  with 
the  axes. 
Resolving  F  parallel  to  each  axis,  we  have  for  the  com- 
ponents 

F.cos.  a,  F  cos.  (3. 

x — a  v — b 

But  cos.  a= and  cos.  (3=- — . 

r  r 

Hence  the  components  are 

r   '        r 

The  other  forces  may  be  resolved  similarly. 
Therefore  the  expression 

Xdx+Ydy=F  i^dx+y-^) 

\    r,  r,     J 

But  r*=(x-ay  +  (y-b)2, 

x — a         v — b 
and,  by  differentiation,  dr— — — dx-\ —  dy. 

c.    .,    ,                           ,        x—a.  .      y—b. 
Similarly,  di\= dx-\ ■ — dy. 

.-.  Xdx+Ydy=Fdr+Fldrl.  ' 
ind  since,  by  hypothesis,  each  fc.Tce  F  is  a  function  of  the  dis- 


212  DYNAMICS. 

tance  r,  each  of  the  terms  of  the  second  member  of  the  equa- 
tion is  an  exact  differential,  and  therefore  the  first  is  also. 

In  the  same  manner,  the  reasoning  may  be  extended  to  any 
number  of  centers  of  force. 

350.  Prop.  To  determine  the  motion  of  a  point  acted  upon  by 
a  force  directed  to  a  fixed  center. 

Assuming  the  fixed  point  as  the  origin  of  co-ordinates,  let  F 
be  the  force,  whether  a  single  one  or  the  resultant  of  several, 
and  r  the  distance  of  any  position  of  the  particle  from  the  cen- 
ter, called  the  radius  vector. 

Multiply  the  first  of  equations  (103)  by  y,  and  the  second 
by  x,  and  subtract  the  former  from  the  latter.     This  gives 

xd"y—yd"x      ,_      __  ,,«„* 

y J       =xY-yH.  (106) 

Resolving  F  in  the  direction  of  the  axis,  we 

have  X=F-,  Y=F^. 

r  r 

Multiplying  the  first  by  y,  and  the  second 

by  x,  and  taking  the  difference,  we  get 

xY-yX=0. 

Therefore  the  first  member  of  (106)  is  zero,  and  since  it  is  the 

xdv — i/dx 
differential  of  — ,  by  integration  we  obtain 


/ 


'xd~y—yd*x    xdy—ydx__ 
W  dt       ~C' 


or                                xdy—ydx=cdt.  (107) 

Let  o)  be  the  angle  made  by  r  with  the  axis  of  x ;  then 

x=r  cos.  o),  (a) 

y=r  sin.  co  ;  (b) 

also,                        dx— cos.  udr—r.  sin.  u.do),  (c) 

dy=s'm.  o)dr+r  cos.  w.cfcd.  (d) 

Multiplying  (c)  by  (6),  and  (d)  by  (a),  and  subtracting,  we 
get 

xdy— ydx = r'dco — cdt. 
r2do) 

•••  -^r-c.  (io8) 


CURVILINEAR     MOTION     OF     A     FREE     POINT.     213 


But  |rWw  is  the  differential  of  the  area  of  a  plane  curve  re- 
ferred to  polar  co-ordinates.  Hence  the  ratio  of  the  element 
of  the  area  to  the  element  of  the  time  is  constant,  and  the  area 
described  by  the  radius  vector  in  any  time  is  proportional  to  the 
time. 

By  integrating  \r^da>=\cdt,  we  have  for  the  area  described 

in  any  time 

k=\ct,  (109) 

and  if  1=1,  we  have  c=2A, 

or  the  constant  c  is  twice  the  area  described  in  the  unit  ot 

time. 

351.  Prop.  Conversely,  if  a  material  point  describe  by  its 
radius  vector  around  a  fixed  center  areas  proportional  to  the 
times,  the  force  is  directed  to  that  point. 

When  the  radius  vector  describes  areas  proportional  to  the 
times,  we  have  (107) 

xdy—ydx 


dt 
Differentiating,  we  get 

xd*y—yd2x 


-c. 


?>J> 


=0. 


dt" 

.  .  (106)  zY-yX=0, 

and  X  :  Y=x  :  y. 

Hence  the  forces  X  and  Y,  which  are  parallel  respectively 
to  the  co-ordinates  x  and  y  of  the  point  P,  are  proportional  to 
these  co-ordinates,  and  the  resultant  PR  of  X  and  Y  must  take 
the  direction  of  PS,  which  is  the  hypothenuse  of  the  right-an- 
gled triangle  PMS,  in  which  PM=y  and  SM=x. 

As  an  application  of  the  general  equations  of  motion  of  a  free 
point  (103),  we  may  take  the  following  cases. 

352.  Prop.  To  determine  the  motion  of  a  point  moving  from 
the  action  of  an  impulsive  force. 

Since,  after  the  action  of  the  impulse,  the  point  is  abandoned 
to  itself,  there  are  no  accelerating  forces  acting  on  the  point, 
and  we  have  (103) 

d*z     „     .    d*y 


dt2 


:X  =  0, 


df 


:Y-t0. 


214  DYNAMICS. 

dx  du 

Integrating,  ^=„lf  ^=«a  ;  (a) 

Uj  and  r2  being  constants,  added  to  complete  the  integral. 

Compounding  these  velocities  (Art.  236),  we  have  for  the 
resultant  velocity 

/dx*+dy9    ds       —— — - 

v=V—de-=dt=Vv°+v" 

The  velocity  is  therefore  uniform. 

Integrating  equations  (a), 

x—v1t+s1,  y—v2t+s2,  (b) 

the  constants  sl  and  s2  being  the  co-ordinates  of  the  point  when 
the  time  commences. 

Eliminating  t  from  equations  (b),  we  have 

ij=-f-x+v2(s2-s,), 
u  i 
the  equation  of  a  straight  line. 

Hence  the  path  of  the  point  is  a  straight  line  and  the  motion 
uniform,  which  accords  with  the  first  law  of  motion. 

353.  Prop.  To  determine  the  motion  of  a  projectile  acted  upon 
by  gravity  regarded  as  a  constant  force. 

In  this  case,  taking  the  axis  of  x  horizontal  and  that  of  y  vei- 
t.ical,  (103)  give 

d2x    _         d7y    ,r 

Multiplying  by  dt,  and  integrating, 

dx  dy 

-7-=!),,  —=v„—gt. 
dt       x     dt       2     * 

Multiplying  by  dt,  and  integrating  again,  we  have 

x=vlt+sl,  y=v2t-lg?+s2.  (a) 

But  if  the  point  of  projection  be  at  the  origin  of  co-ordinates, 
and  t  be  reckoned  from  the  commencement  of  motion,  s,  ana  sa 
will  each  be  zero.  Putting  v2=avx,  s1=0,  s2  =  0,  and  elimi- 
nating t  from  the  equations  (a),  we  get 

.  g 
y=ax—x—-2-. 

Compare  this  with  (63). 


CONSTRAINED     MOTION     OF     A     POINT. 


215 


§  III.    CONSTRAINED    MOTION    OF    A    POINT. 

354.  Prop.  To  determine  the  velocity  of  a  point  moving  on  a 
given  curve. 

If  a  point  is  constrained  to  move  on  a  curve,  the  reaction  of 
:he  curve  will  be  a  new  force  normal  to  the  curve,  and  may 
be  resolved  in  the  direction  of  the  axes  and  combined  with  the 
other  components,  as  in  Statics,  Art.  77. 

Let  N  be  this  normal  force,  a  and  3  the  angles  which  the 
normal  makes  with  the  axes,  and  X  and  Y  the  sums  of  the 
components  of  all  the  other  forces  in  the  direction  of  the  axes. 

Resolving  N  in  the  same  directions,  we  shall  have  from  (103) 
d'x 


di 


-=X+N  cos.  a 


§=Y+Ncos.0 

But  cos.  a—  sin.  rtm=+-r-,  supposing 
the  motion  in  the  direction  tV,  and  cos. 
8=  sin.  a- 


N  M 


_  dx 

cos.  rtm= — — . 
ds 


lf-x+NTs 

de  ds' 


Multiplying  (a)  by  2dx,  and  (b)  by  2dy,  and  adding 
2dxd7x+2dyd*y 


de 


=  2(Xdx+Ydij). 


The  first  member  is  the  differential  of 
dx*+dy*_ds*_  a 
df  ~~df~V' 
Whence,  integrating, 

v9=2f(Xdx+Ydy)+c.  (110) 

Cor.  1.  Since  this  result  is  exactly  the  same  as  that  obtained 
for  unrestrained  motion  (105),  we  may  conclude. 


216  DYNAMICS. 

1°.  That,  if  no  accelerating  force  act  on  the  point,  or  X=0 
and  Y=0,  its  velocity  will  remain  constant,  and  be  not  at  all 
retarded  by  the  action  of  the  curve.     See  Art.  284. 

2°.  If  any  accelerating  forces  do  act  on  the  point,  the  veloc 
ity  is  independent  of  the  curve  on  which  it  is  constrained  to 
move.     See  Art.  286. 

Cor.  2.  If  the  accelerating  forces  are  all  parallel,  they  may 
be  assumed  parallel  to  the  axis  of  x,  and  in  this  case  (110)  be 
comes 

v*=2fXdx+c.  (Ill) 

Cor.  3.  If  in  this  last  case  the  force,  or  resultant  of  the  forces, 
is  constant,  and  equal  to/,  we  have 

u2=±2fx+c, 
where  the  upper  or  lower  sign  is  to  be  used  according  as  the 
force  tends  to  increase  or  diminish  x. 

If  the  distance  of  the  point  from  .the  origin,  x—a,  when  v=0. 
c=2fa,  and 

v-=2f(a-x);  (112) 

and  since  only  the  ordinates  on  the  axis  of  x  are  involved,  the 
velocity  of  the  point  on  the  curve  depends  not  on  the  curve  de- 
scribed, but  on  the  difference  of  the  ordinates  on  the  axis  of  a:. 

355.  Prop.   To  find  the  time  of  motion  of  a  point  on  a  given 

curve. 

ds 
In  all  cases  [VII.],  dt=— . 

Hence,  when  the  nature  of  the  curve  And  the  velocity  at  any 

point  of  it  is  known,  the  value  of  v  being  found  from  (110)  and 

substituted  in  that  of  dt,  the  time  may  be  found  by  integration 

Cor.  If  the  forces  act  in  parallel  lines,  and  their  resultant  is 

constant,  by  (112) 

ds 
dt=  =.  (113) 

V2f(a-x) 

356.  Prop.   To  find  the  reaction  of  the  curve. 

dxi  dx 

Multiplying  (a),  Art.  354,  by  -j-,  and  (b)  by  -=-,  anc  subtract 

iog  the  latter  from  the  former,  we  get 


CONSTRAINED     MOTION     OF     A      POINT.  217 

N— Y— -  X—  4-  'lyd  *X  ~  (/"'d  '' 


'  ds        ds  dsdt" 

t,,.    .       .        ,  ,       ,  .  ds 

Eliminating  dt  by  the  equation.  v= -r-, 

dx        dy      a  dyd2x  —  dxd~y 
ds        ds  ds3 

But  if  p  be  the  radius  of  the  osculating  circle, 

ds5 
P 


dyd^x—dxd^y 
__     ^Tdx     vdy     v" 

ds        ds     p 

v\ 
=  Y  cos.  (3— X  cos.  a-\ — 
P 

The  first  two  terms  of  this  value  of  N  are  equal  and  oppo- 
site to  the  forces  X  and  Y  resolved  in  the  direction  of  the  nor- 
mal.    They  give,  therefore,  the  pressure  on  the  curve  due  tc 

v~ 
the  action  of  these  forces,  while  the  other  term  —  is  the  reac- 

P 

tion  of  the  curve  or  pressure  due  to  the  motion. 

Otherwise.  If  we  suppose  no  accelerating  forces  to  act  on 
the  point,  then  X  and  Y  are  each  equal  to  zero  in  (a)  and  (b), 
Art.  354.     Hence 

N  Jdx'+dy'\  =N,Jd*xy  +  (d*yy 


dt' 


ds 

or,  eliminating  dt  by  the  equation  dt= — , 


N=B  • 5? • 

But  the  coefficient  of  u2  is  the  reciprocal  of  the  radius  of 
curvature  when  s  is  the  independent  variable. 

.-.  N=-  (114) 

Compare  this  with  (79). 

As  an  application  of  the  formulae  of  constrained  motion,  we 
may  take  the  following  rases  : 


218  DYNAMICS. 

357.  Prop.  To  determine  the  motion  of  a  body  descending  by 
the  force  of  gravity  down  the  arc  of  a  vertical  circle. 

1°.  To  find  the  velocity.  Formula  (112)  applies  to  this  case, 
and  if  A=SA  be  the  height  from  which  the 
body  at  P  descends,  and  x  any  distance  from 
the  lowest  point  on  SA,  the  axis  of  x,  we  have 

v=  V2g(h—x). 

When  x=0,  we  have  for  the  velocity  at  the  lowest  point, 

u=  V2gh, 

which  is  the  same  as  that  due  to  the  vertical  height  h. 

2°.  To  find  the  time,  we  have  from  (113) 

ds 
dt=- 


V2g(h—x) 
The  equation  of  the  circle  is 

y*=2ax— x2. 
Hence  _(*-*)  W 

But  ds°-=dx2+dy2=dx\l  + 


V2ax—x'1 

This,  taken  negative,  because  the  arc  is  a  decreasing  fur.c 
tion  of  the  time,  and  substituted  in  the  above,  gives 
a  dx 


dt=  — 


V  2g  V  (h — x)  (2ax — x2) 
a  dx 


V2g  V(hx-x*)  (2a-x) 

a     (2a—x)~2dx 
V2g     Vhx—x'1 


(«) 


—  *J± 


"  v       2a  J 


g      Vhx- 


CONSTRAINED     MOTION     OF     A     POINT.  219 

Expanding  (1  —  — -  J    by  the  binomial  theorem,  we  have 
1     fa       dx       ,       \(x\     \Z(x\*    lZ5(x\% 

&c.}. 

— x"dx 


Thus  the  terms  to  be  integrated  are  of  the  form 

Vhx— x1, 

the  exponents  n  being  natural  numbers  beginning  with  0.  Per- 
forming the  operations  and  taking  the  integral  between  the 
limits  x=h  and  x—Q,  we  find 

Hsv/f  * l + G) '•  (s) + GD 3  GD 2+ Gil) '  (£)  '+* 

&c.|. 

If /i  be  small,  the  first  term  will  give  an  approximate  value 
for  the  time,  viz.. 

Otherwise.  If  in  (a)  #  be  rejected  as  small  in  comparison 
with  2a,  the  equation  reduces  to 

i     fa  _J^__ 

dt=-W~.    .-j- ,. 

v  g-  Vnx—x 

.     /a  .       2x 

.'.  *=-£y-.versin.  l^-+c. 

But  2=0  when  x=h; 

and  ^=^-\/— (tt— versin._1-r- ), 

and  when  a;=0,        £=£tt\/-.     (See  (8G) ). 

© 

358.  Prop.  To  determine  the  motion  of  a  body  descending  by 
gravity  down  the  inverted  arc  of  a  cycloid  whose  base  is  hori- 
zontal and  axis  vertical. 

The  cycloid  being  inverted,  let  the  origin  be  at  the  lowest 


220 


])  Y  K  A  M  1  C 


point,  the  axis  of  x  being  vertical,  and  that  of  y  horizontal 
Let  the  axis  AB  =  2a,  AM=.r,  PM=?/,  and  AP=s. 

The  velocity,  as  in  the  preceding  propo* 
~J     sition,  is 


t;=  V2g(h—x), 

h  being  the  abscissa  of  the  point  K  from 

which  the  boay  begins  to  descend. 

The  differential  equation  of  the  curve,  referred  to  the  vertex 

as  the  origin,  is 

(2ax  — x)dx 
dy= — , 
9       V2ax-x'2 

/(2a— xY 
But  ds* = dtf + dx*  =  (  ^ -s + 1 ) dx\ 

.'.  ds=dx\/ — . 
V    x 

Substituting  this  value  of  ds  with  its  proper  sign  in  [VII.], 

we  find 

/a~      dx 

V  g'  Vhx—x2' 

Integrating,  t=  y  — .  (tt— versin.-1-r- ) , 

and  therefore  for  the  time  of  descent  to  the  lowest  point,  where 
x=0, 

/a     tt     /4a  ,  ,   . 

t=n\/-= ba/— •  (115) 

V  g    2  V   g 

Cor.  The  time  t  being  independent  of  the  abscissa  h,  will  be 
the  same,  from  whatever  point  in  the  curve  the  body  begins  to 
descend  ;  in  this  respect  differing  from  the  motion  in  circular 
arcs.  On  account  of  this  remarkable  property  the  cycloid  is 
called  tautochronal. 

359.  Schol.  The  property  of  taatochronism  which  attaches 
to  the  cycloid  was  supposed  to  give  peculiar  advantages  to  a 
pendulum  made  to  oscillate  in  this  curve.  The  method  of  ac- 
complishing this  is  naturally  suggested  by  the  property  of  the 
evolutes ;  and  as  an  evolute  of  a  cycloid  is  another  equal  cy- 


CENTRAL     FOKCE.  22 1 

cloid,  wo  have  only  to  take  two  equal  semicycloids  and  place 

the  extremities  of  their  bases  contiguous  in 

s 

the  same  horizontal  line  as  in  the  figure. 
If  we  then  suspend  a  body  at  the  point  S, 
common  to  the  two  semicycloids,  by  a  flexi- 
ble string  equal  in  length  to  either  semicy- 
cloid,  or,  which  is  the  same  thing,  to  twice 
the  axis,  and  make  it  oscillate  between  the  two,  it  will  generate 
the  involute,  or  another  equal  cycloid. 

The  constant  change,  however,  in  the  center  of  motion 
arising  from  the  contact  of  the  string  with  the  two  curves  or 
checks,  which  changes  also  the  relative  velocity  of  the  different 
parts  of  the  vibrating  body,  renders  this  contrivance,  although 
beautiful  in  theory,  yet  useless  in  practice ;  independently  of 
the  difficulty  of  obtaining  a  string  sufficiently  flexible,  and  of 
ensuring  accuracy  in  the  plates. 

Vibrations,  therefore,  in  small  circular  arcs,  which  are  at  the 
same  time  also  most  natural,  have  been  adhered  to  in  practice, 
and  it  has  been  shown  {Art.  302  and  300)  that,  as  long  as  the 
arc  is  small,  these  vibrations  have  all  the  advantages  of  the 
vibrations  in  cycloidal  arcs. 


§  IV.    MOTION    OF    A   POINT   ACTED   UPON   BY   A   CENTRAL   FORCE 

360.  The  case  of  a  point  revolving  in  an  orbit  by  the  action 
of  a  force  tending  to  a  fixed  center  is  of  sufficient  importance 
to  justify  a  distinct  discussion,  especially  as  the  formulae  are 
susceptible  of  considerable  simplification,  when  the  force  is 
given  as  a  function  of  the  distance  from  the  center.  The  fun 
damental  equations 

A-~dt"   X~df" 

are  components  of  the  force  in  two  rectangular  directions 
But  when  the  force  is  directed  to  or  from  a  fixed  center,  and 
is  given  in  functions  of  the  distance  of  the  particle  from  that 
center,  it  is  most  natural  and  convenient  to  introduce  polar  co- 
ordinates. 


222  DYNAMICS. 

361.  Prop.  To  transform  the  expressions  foi  the  force  from, 
rectangular  to  polar  co-ordinates. 

Suppose  the  origin  to  be  at  S,  the  center  of 
force,  and  let  7'=SP,  the  radius  vector  or  dis- 
y^^\       tance  of  the  particle,  and  w=PSX,  the  angle 
made  by  r  with  axis  of  x. 

Then     X=  —  F  cos.  w,  Y=—  F  sin.  u, 
the  signs  being  negative  because  the  force  is  supposed  to  be 
attractive,  or  act  toward  the  center  in  opposition  to  X  and  Y. 
Multiplying  the  first  by  sin.  w,  and  the  second  by  cos.  w,  and 
subtracting,  we  have 

X  sin.  w— Y  cos.  u=0.  (a) 

Similarly,  multiplying  the  first  by  cos.  w,  and  the  second  b\ 
sin.  o),  and  adding,  we  get 

X  cos.  w+Y  sin.  «— — F.  (b) 

But  x  and  y  being  the  co-ordinates  of  the  position  of  the 
particle 

x=r  cos.  o),  y=r  sin.  w. 
Differentiating  each  twice,  we  obtain 

d*x=d*r.cos.  (o—2dr.do).s\n.  cj—rdo^.cos.oi—rd'^.s'in.  w, 
d^y^d^r.  sin.  u+2dr.du.cos.  a— rd^.s'm.  (o+rcPw.cos.  <y. 

By  dividing  these  expressions  by  df,  the  values  of  X  and  Y 
will  be  obtained  (103),  and  if  we  substitute  these  values  in  (a) 
and  (b),  we  get 

drdo)       d2o> 

which  are  two  polar  equations  to  determine  the  motion. 

362.  Cor.  I.  The  radius  vector  of  a  point  revolving  about  a 
center  of  force  describes  equal  areas  in  equal  times. 

For,  multiplying  (116)  by  rdt,  we  obtain 
,   d<*>      „d2(o  ,rWw 


t 

CENTRAL     FORCES.  223 

Integrating,  ~dT=Ci  (118) 

and  fr'd(o=ct-\-ci. 

But  \fr~du  is  the  area  described  by  the  radius  vector  in  the 
time  t,  and  it  varies  as  t. 

363.  Cor.  2.   The  angular  velocity  of  the  radius  vector  varies 
inversely  as  the  square  of  the  distance  of  the  particle. 

„  d<*>     c 

From  (118)  we  obtain       -77=-.  (119) 


dt     r 

i<3    the*  nnmilnr   vpI 

dt 


But  -j7  is  the  angular  velocity  of  r,  and  it  varies  inversely 


as  r . 

364.  Prop.  To  determine  the  velocity  of  the  -particle  in  its 
orbit. 

Multiply  (116)  by  rdu,  and  (117)  by  dr,  and  add  the  results. 
This  gives 

■    d'r       7  dtf      57    d2o>    ^7 
,  fdr2       rfws\        _ , 

d\-d7+>^)+2Fdr=°- 

Therefore,  integrating,  we  have 
dr2+r'du 


df 


+2fFdr=c'.  (120) 

.     ds2    rfra+rW        a  3* 

But  de  =  — dt* — *    (  367'  4°^ 

..  v2=c'-2fFdr.  (121) 

Cor.  Since  the  value  of  v  depends  only  on  the  distance  01 
the  particle  from  the  center,  the  velocity  of  the  particle  will  be 
the  same  at  any  two  points  equally  distant  from  the  center,  and 
the  velocity  acquired  in  passing  from  one  point  to  another  will 
be  the  same,  whatever  be  the  path  or  curve  between  them,  if 
the  law  of  force  remain  the  same. 

365.  Prop.  To  find  the  time  of  describing  any  portion  of  the 
orbit. 


224 


DYNAMICS. 


By  (119), 


which,  substituted  in  (120),  gives 


dr*     c2 
-+p+2/F<fr=c'. 


(122) 


In  this  expression  we  have  the  relation  of  r  to  t,  independent 
of  the  orbit. 


366.  Prop.   To  find  the  equation  of  the  orbit. 

r^du? 
Substituting  in  (122)  for  dt"  its  value  — — , 


obtained  from 


(118),  we  get 


/  dr2       1  \ 


or,  to  give  the  equation  a  more  convenient  form,  put  r=-,  from 


which  we  find  dr= 
in  (123),  give 


du 


These  values  of  r  and  dr,  substituted 


c 


du2       „        ^du 
~+u-)-2/F^=c>.  (124) 


When  the  law  of  force  is  given,  we  can  obtain  from  this  ex- 
pression, by  integration,  the  relation  between  u  and  w,  or  the 
polar  equation  of  the  orbit ;  or,  if  the  relation  between  u  and  &> 
is  given,  the  law  of  force  may  be  determined. 

367.  Schol.  A  familiarity  with  the  geometrical  representa- 
tives of  the  terms  and  factors  in  the  foregoing  formula?  will 
conduce  to  facility  in  their  application. 

In  the  annexed  figure  let  the 
center  of  force  be  at  S,  the  origin 
of  co-ordinates,  SP=r  the  radius 
vector  of  the  particle  at  P,  W=ds 
x   an  element  of  its  path  coinciding 
A  s  "  with  the  tangent  PT,  w=PST  the 

angle  made  by  the  radius  vector  with  the  axis  of  x,  P/SP=c/oj 
the  angle  described  by  the  radius  vector  in  the  indefinitely 


CENTRAL     FORCES.  225 

small  time  dt,  and  mV  =dr  the  increment  or  decrement  of  the 
radius  vector  in  the  same  time.  Let  the  arc  Vm  be  described 
with  S  as  a  center,  and  radius  SP,  and  the  arc  nn'  with  the 
radius  Sn=l.     Then, 

1°.  Since  Sn  :  SP=wn'  :  Vm  or  1  :  r=doj  :  Vm, 

Vm=rdw. 
2°.  The  area  of  the  sector  SPm  is 

A= — - — =h-*do). 
2  2 

3°.  If  the  point  P'  be  indefinitely  near  to  P,  PP'  and  Vm  may 

be  considered  straight  lines,  and  the  triangle  PmP'  a  plane 

rectilinear  triangle,  right-angled  at  m.     Hence 

P'P^Pm'+P'm2, 

ds*=r*du?-\-dri. 

4°.  If  Sp—p  be  a  perpendicular  on  the  tangent,  when  the 

triangle  PwzP'  is  in  its  nascent  or  evanescent  state,  the  angle 

PP'm=^PS  and  the  triangle  PP'm  is  similar  to  joPS.     Hence 

PP  :  Pm=PS  :  Sp,  or  ds  :  rda=r  :  p.     .:  P=^,  and  [VII.], 

r*doi  ?-"d(o  . 

/>= — =-.     .*.  vp——rr  =  ,  by  (118),  twice  the  area  described 

in  the  unit  of  time. 

?Jd(i)'        r^dui2 


A1S°'  P*-'ds>  ~dr*+rW 

5°.  Since  S/j  :  pV  :  PS=Pm  :  m?'  :  V'V=rdo)  :  dr  :  ds, 
Sp _Vm     rd(ji_  . 
•*'  SP^PP^Is""8111,  F» 

PS~PP~  rfs  ~C0S-  F' 

»S     raP      rc?w 
and  — s=— T57=-j-=tan.  P. 

pr    mr       ar 

6°.  If  the  force  F  be  resolved  in  the  direction  of  the  curve, 

«re  shall  have 

d*s        ^  „        „di- 

=  _p  Cos.  P^-F^. 

dC  ds 


£26  DYNAMICS. 

Multiplying  both  sides  by  2ds,  and  integrating,  we  have 

ds* 

—=v*=c'-2fFdr,  as  before  (121). 

By  reference  to  (117),  we  see  that  the  first  term  has  the 

form  of  the  differential  expression  [IX.]  of  a  force  in  the  direc- 

c         ,  ,  rdbf     r'do)-      __      rdo)  .     .  e 

tion  oi  v ;  the  second  term  —rz-= — tt~-     But —7-  is  the  arc  01 

at        rdt  dt 

a  circle  of  radius  r,  divided  by  the  time.     It  is,  therefore,  the 

linear  velocity  d,  of  a  body  in  a  circle.     Hence     7  „  =—=f. 

J     1  J  dt        r    - 

the  centrifugal  force  in  a  circle.     .*.  (117)  becomes  -7-7-=/— F, 

F  being  the  force  by  which  the  point  describes  any  orbit,' and 
/  the  centrifugal  force  in  a  circle  whose  radius  is  r,  the  dis- 

tance  of  the  point  at  the  instant,     -jj,  the  difference  of  these 

forces  at  any  time,  is  that  by  which  the  radius  vector  is  in- 
creased or  diminished,  and  is  called  the  paracentric  force.     In 

d*r        dr 
like  manner,  the  integral  of  -7-7  or  -=-  is  the  velocity  of  ap 

proach  or  recession  from  the  center,  and  is  called  the  para 
centric  velocity.  The  paracentric  force  is  the  difference  be 
tween  the  centrifugal  and  centripetal  forces. 

rdui2     1  r4e?«s     ca 


Cor.  1.  Since 


df      r3*  dt2      r3 


.'.  /=£  (125) 

Cor.  2.  The  quantity  by  which  c~  is  multiplied  in  (123)  is, 
by  (4°),  equal  to  — j.     Hence 

-=+2fFdr=&. 

P      * 

Differentiating,  p  being  variable, 

2cVd       „ , 
J-+2Fdr=0. 


CENTRAL     FORCES.  227 

Hence  F=^r>  (120) 

p  dr  v 

an  expression  most  useful  in  finding  the  law  of  force  by  whict 
a  given  orbit  may  be  described. 

/-.  o-  ■        2     ds2     dr+r\lu?     r'du2     e2 

Lor.  3.  Since  v  =-^=- 7^ —    „  ,,■>=—; 

dt  dt  p'dt"    p~ 

c2     Fpdr       „  pdr  „       _  pdr 

and  (126),  -=^-=2F.^,  ,.  »'=2F^-.        (127, 

pdr 
But  77-r -=i  chord  of  curvature  through  the  pole,  and  (if  F 
2dp     *  o  l  \ 

be  constant)  is  (55)  the  space  through  which  a  body  must  fall 
to  acquire  the  velocity  v.  Hence  the  velocity  of  a  body  at  any 
point  of  its  orbit  is  equal  to  that  acquired  by  falling  through 
one  fourth  the  chord  of  curvature  through  the  pole. 

Application  of  the  preceding  formulae  to  the  motions  of  the 
planets. 

368.  Observation  has  established  three  facts  respecting  the 
motions  of  the  planets,  which,  from  their  discoverer,  are  called 
Kepler's  Laws. 

1°.  The  areas  described  by  the  radius  vector  of  a  planet  art 
proportional  to  the  times. 

2°.  The  orbit  of  a  planet  is  an  ellipse  of  which  the  center  of 
the  sun  is  one  of  the  foci. 

3°.  The  squares  of  the  times  of  revolution  of  the  different 
planets  are  proportional  to  the  cubes  of  their  mean  distances 
from  the  sun,  or  the  semi-major  axes  of  their  orbits. 

These  laws  relate  only  to  the  center  of  inertia  of  each  planet, 
and  conclusions  from  them  must  be  limited  to  the  motion  of 
that  point;  in  other  words,  to  the  motion  of  translation  of  the 
planets. 

3G9.  Prop.  The  accelerating  force  by  which  a  planet  describes 
its  orbit  is  directed  to  the  center  of  the  sun. 

For,  by  the  first  of  Kepler's  laws,  the  areas  described  by 
the  radius  vector  are  proportional  to  the  times,  and  when  this 
is  the  case,  by  Art.  351,  the  force  must  be  directed  to  the  point 
about  which  this  equable  description  of  areas  takes  place. 


228  DYNAMICS. 

370.  Prop.   The  force  by  which  a  planet  describes  its  orbit 
varies  inversely  as  the  square  of  its  distance  from  the  center. 

By  the  second  law  the  orbit  is  an  ellipse.     The  equation  of 
an  ellipse  referred  to  the  focus  as  a  pole  is. 

ail-e2) 
1+e  cos.  (to— 6)  v       ' 

in  which  a  is  the  semi-major  axis,  called  the  mean  distance,  and 
e  the  eccentricity.  The  angle  6  is  an  angle 
made  by  a  fixed  line,  through  the  focus  with 
the  major  axis,  reckoned  from  the  lower 
apsis,  or  least  distance,  and  called  the  longi- 
tude of  the  perihelion,  the  lower  apsis  being  called  the  perihel- 
ion, and  the  higher  the  aphelion.  The  angle  w— 6,  which  ex- 
presses the  angular  distance  from  the  perihelion,  is  called  the 
true  anomaly. 

Putting  r=-  in  the  above  equation,  we  get 

1+e  cos.  (o— 6)  ■         .  . 

U==       a(l-e>)       '  <& 

Differentiating  in  reference  to  w, 

du_     e.sin.  (to— 6) 
dto  a(l—e2) 

Differentiating,  again, 

d'u        e  cos.  (to— 6) 


dto'  a(l-e') 

Adding  (b)  to  (a), 

d'u  1 


(P) 


dto2  a(\—e)                            y  ' 
Differentiating  (124), 

d'u  F 

dto'  cu 

Hence  F=^-r=^-}}?.  (129) 

The  coefficient  of  —  being  constant,  the  force  varies  inverse 
r 

Iv  as  the  square  of  the  distance. 


CENTRAL     FORCES.  229 

Ifr=l,  F=~7i i\- 

a(l—e) 

Hence  the  coefficient  of  —2  is  the  intensity  of  the  force  at  the 
unit  of  distance,  or  the  absolute  force. 

371.  Prop.  The  intensity  of  the  force  is  the  same  for  all  the 
planets  at  the  same  distance. 

The  quantity  c  (109)  is  twice  the  area  described  in  the  unit, 
of  time.  If,  therefore,  T  be  the  time  of  one  revolution  of  a 
planet,  cT  will  be  twice  the  area  of  the  ellipse  described  by 
the  radius  vector. 

But  the  area  of  an  ellipse  is  7rab='na"  Vl—  e\ 

.-.  cT  =  2rraWl-e\ 
c       _4ttV 
and  a(l-ea)  =  ~TT~- 

Similarly,  if  a',  e',  c',  T'  express  the  same  quantities  in  any 
other  orbit,  we  shall  have 

c'2        _4ttV3 
a'(l-e'2)~   T'2  * 
By  Kepler's  third  law, 

T  :  T/2  =  «3  :  a'\ 
c2  c'2 


'     a(l-e2)     a'(\-e'"-)' 

Hence  the  force  does  not  vary  from  one  planet  to  another 
except  in  consequence  of  change  of  distance. 

372.  Prop.   To  find  file  velocity  of  a  planet  at  any  point  in  its 
vrbit. 

By  (121),  v2=c'-2fFdr. 

Substituting  in  this  the  value  of  F  (129),  we  have 
»-  2c2         rdr 

V  ~C'~~a(l-e2)V   V' 
2c2       1 
Integrating,  v  =c+an_e*y? 

To  determine  c',  we  must  know  the  velocity  at  some  given 


230 


DYNAMICS. 


distance,  or  at  what  distance  and  with  what  velocity  the  planet 

was  originally  projected  into  space.  ' 

Let  Uj  be  this  primitive  velocity,  and  ;-,  the  corresponding 

distance. 

2c2        1 
Then  v\=c'+- 


or  c'=v\ 


2c"        1 


1     ^l-e2)';-,' 

,       .         2c2       (\      1  \  ,      x 

•*• v  =v i+  n    »x-( — ;■     (13°) 

a{\  —  e)  \r     r,'  v 

Otherwise.  From  Art.  367,  Cor.  3,  we  have 

c" 

and  (Analytic  Geometry)    j5"-— 


2<2— r       2a— r 
c2        2a —r 


(131) 


<r(l-e2)'     r 

Cor.  1.  Hence  the  velocity  is  greatest  when  r  is  least  or  al 

perihelion,  and  least  at  aphelion,  where  r  is  the  greatest. 

Cor.  2.  If  a  body  move  in  a  circle  wThose  radius  is  r,  vl  be 

ing  its  velocity,  the  central  force  is 

v\  c        1 

F=i=a(l-e*)'?  (129)' 

if  the  force  is  the  same  as  that  which  retains  the  planet  in  its  orbit 

c2        2a— r         c-       1 

.*.  v  '.  v== . : .— 

1     cr(l  —  e")      r         a(\  —  e")'i 

=  2a  —  r  :  a. 

That  is,  the  square  of  the  velocity  in  the  ellipse  is  to  the  square 

of  the  velocity  in  the  circle  as  the  distance  of  the  planet  from 

the  unoccupied  focus  is  to  the  semi-major  axis. 

Cor.  3.  If  7-,  =  perihelion  distance,  and  r2  =  aphelion  distance 

c2        2d  —  r  c2        r 

then,  if  in  (131)  r=rx,  v*=-^- jr. l=in-- 57.— ■     («) 

a(l  —  e)      r,  a  (\—e)  rx 

„_       c2        2a— r2_       c2        r, 
r-ra,  v  -a^x__ey     ,*    ~aXl-e*)7^'     (  ' 


r==a'  ■Ps=5i(i=5rBt/w^,i 


CENTRAL     FORCES.  231 

or,  the  velocity  at  the  extremity  of  the  minor  axis  is  a  mean 
proportional  between  the  velocities  at  the  perihelion  and 
aphelion. 

373.  Prcp.  To  determine  the  orbit  which  a  body  will  describe 
when  the  force  varies  inversely  as  the  square  of  the  distance. 

If  ([*  be  the  force  at  the  unit  of  distance,  then 

.'.  fF^=ffidu=fiu. 
This,  substituted  in  (124),  gives 


(dur 


which  is  the  differential  equation  of  the  orbit. 
To  integrate  this,  we  find 

c^du? 


do)"- 


2„.2* 


c'+2fiu— c"*u 


(i  1 

Put  u=z+— ,  and  b2=— (c'«r+M"),  and  we  have 
c  c 

dz 
doi- 


Vb2-z* 

_z 
or,  integrating,  «=0+cos.    -. 

.'.  z=b  cos.  (to  —  6) ; 
or,  replacing  the  value  of  z, 

u—-^+b  cos.  (to— 6). 

cl 

Hence  r= j-j— ,  (132) 

H ■  cos.  (to  —  6) 

which  is  the  equation  of  a  conic  section  referred  to  the  focus*. 
Hence  the  orbit  which  a  body  will  describe  when  the  force 
varies  inversely  as  the  square  of  the  distance  is  a  conic  section. 
To  determine  the  constants,  and  therefore  the  dimensions 
and  form  of  the  orbit,  some  other  circumstances  must  be  giveiL 


232  DYNAMICS. 

Suppose  we  know  vx  the  initial  velocity  or  velocity  of  pro- 
jection into  space,  p  the  distance  of  the  planet  from  the  sun, 
and  ij)  the  angle  made  by  the  direction  of  projection  with  the 
distance. 

By  Art.  367,  5°,  sin.  V=^A 

as 

it  /     ^\  P^du       ds  ado 

Hence  (118),     c=—=p-— =p.„.  Sin.^.  (a) 

By  (121),  v*=c'-2/Fdr=c'+—. 

Comparing  (132)  with  (128),  we  see  that 

c2  6e2     1  i 

a(l—  e')=—  and  e= — =-(c'c2+u2V. 


c'     2(i—v\p 


(0 


and  e2=^-+i  =  i+-  _ ^_L,     (</) 

Cor.  From  (c)  it  appears  that  the  semi-major  axis  a  depends 
only  on  the  distance  p  and  the  velocity  of  projection  i>,,  and 
is  independent  of  the  angle  of  projection  ip.  Hence,  in  what- 
ever direction  the  body  is  projected,  the  major  axis  of  the  orbit 
will  be  the  same. 

The  orbit  will  be  an  ellipse,  hyperbola,  or  parabola,  accord- 
ing as  the  value  of  e  is  less  than,  greater  than,  or  equal  to  unity, 
or  (d)  according  as 

>s  negative,  positive,  or  equal  to  zero. 

Or,  since  (99),  v*=  ^(1--),  if  a=cc,  v2=— .     Therefore 
x  x        a/  x 

when  a  body  is  attracted  from  an  infinite  distance  to  a  center 

of  force  which  varies  inversely  as  the  square  of  the  distance 

the  square  of  the  velocity  at  any  distance  p  is 


CENTRAL     FORCES.  23li 

n        2W  .        _         2/4 

v;  = —  and  v. =0. 

P  P 

Hence  the  orbit  will  be  an  ellipse,  parabola,  or  hyperbola,  ac- 
cording as  the  velocity  of  projection  is  less  than,  equal  to.  01 
greater  than,  that  acquired  from  an  infinite  distance. 

374.  Prop.   To  find  the  time  of  describing  any  portion  of  th? 
orbit. 

Substituting  the  value  of  F=—  in  (122),  we  get 

dr2     ca     2[i_ 

df  '  V~~~C' 

The  values  of  the  constants  found  in  terms  of  the  elements 

of  the  orbit  in  the  preceding  proposition  are 

fi 
c'= —  and  c'=ua(l—  e~). 
a 

Substituting  these  values,  and  multiplying  by  — ,  wc  obtain 

ar*dr*  v,      „. 

£-5r+.(«-r)-a«=0. 

a\2  rdr 


Hence  dt=  . 

#)     vVe2-(a-r)2 

In  order  to  integrate  this  expression,  let  {a— r).=aez,  then 

i 
/a.\  2  (1—  ez)dz 
dt=—a' 


V-J      Vl-z2 
a3  \  2   (      —  dz  zdz      j 


P'    '  I  Vl-z2       Vl-r  \  ' 
Integrating,      t=[—J   {cos.  lz— e(l— za)3]+k. 

When  z=l,  or  r=a—ae,  and  the  body  is  at  the  nearer  apsis, 
then  tl=k=  the  time  at  the  perihelion. 

Hence  the  time  through  any  portion  of  the  arc  from  the 
perihelion  is 

t"t^=  (7)  3{C0S "1*-e<1  "Z^'  •      (133) 


234  I)  Y  N  A  M  I  C  S. 

If  z=  —  l,r=a+ae»  and  the  body  is  at  the  further  apsis,  in 
which  case  t—ti  is  the  time  of  half  a  revolution,  and 

fa* 

l-t     =71  I    - 

1     \/* 

Also,  if  T  be  the  periodic  lime, 

Ta- *a\ 

I" 

or  the  square  of  the  periodic  time  varies  as  the  cube  of  the 
mean  distance. 

375.  Schol.  The  quantity  \i  which  enters  into  the  results  of 
the  preceding  investigations  is  the  value  of  the  accelerating 
force  at  the  unit  of  distance  from  the  center  of  force.  Now 
the  attractive  forces  of  the  sun  and  planets  vary  directly  as 
their  masses,  and  if  M  be  the  number  of  units  of  mass  of  the 
sun,  and  in  the  same  of  any  planet,  and  if  we  assume  for  the 
unit  of  force  the  attraction  of  a  unit  of  mass  at  a  unit  of  dis- 
tance, M  will  express  the  attractive  force  of  the  sun  at  the  unit 
of  distance,  and  m  that  of  any  planet ;  and  the  whole  force  by 
which  they  will  tend  to  approach  each  other,  or  the  whole  force 
which  the  sun,  regarded  as  fixed,  exerts  on  the  planet  at  the 
unit  of  distance  is  M+m=/i ;  and   for  any  other  distance  r 

M+m_fi 

The  intensity  of  the  solar  and  planetary  attractions  may  be 
expressed  in  terms  of  terrestrial  gravity.  For  this,  let  r,  be 
the  radius  of  the  earth,  and  mx  its  mass,  and  since  the  forces 
are  directly  as  the  masses,  and  inversely  as  the  squares  of  the 
distances,  we  have 

m,     M_       Mr; 

~r\  :  ~7~S  '  m^Pg' 

the  fourth  term  being  the  attractive  force  of  the  sun  at  any  dis- 

TIXV 

tnnce  r.     Similarly,  the  attractive  force  of  the  planet  is l-g: 

J  l  m^ 

.   n  .     M+jfi  r\        fi  r\ 

hence  their  joint  influence   is   — = — . — g=~i- — g,  in  whicli 
J  r       ?n°     r  m° 


CENTRAL     FORCES.  235 

— L.iT  is  that  which  was  assumed  above  for  the  unit  of  force,  01 
the  attraction  of  the  unit  of  mass  at  the  unit  of  distance. 

376.    EXAMPLES. 

Ex.  1.  Required  the  velocity  and  time  when  a  materia] 
particle  is  attracted  to  a  fixed  point  by  a  force  varying  in- 
versely as  the  square  root  of  the  distance. 

II  -LI 

Ans.  v=2fis(a3—x2)2 
2 

t=^J^+2ah  i^-x^f. 

Ex.  2  Find  the  velocity  and  periodic  time  of  a  body  re- 
volving in  a  circle  at  a  distance  of  n  radii  from  the  earth's 
center. 

Ans.  v=  (£. 
\  n 

Ex.  3.  Find  the  least  velocity  with  which  a  body  must  be 
projected  from  the  moon  in  the  direction  of  a  line  joining  the 
center  of  the  earth  and  moon,  so  that  it  may  reach  the  earth. 

Ex.  A.  Given  the  velocity,  distance,  and  direction  of  pro- 
jection, when  the  force  is  attractive  and  varies  as  the  distance, 
to  find  the  orbit. 

-  3 
Ex.  5.  A  body  is.  projected  in  a  direction  which  makes  an  c 

angle  of  60°  with  the  distance,  with  a  velocity  which  is  to  the     n 

velocity  from  infinity  as  1  :  VS  :  the  force  varying  inversely 

as  the  square  of  the  distance.     Find  the  major-axis,  the  posi-  - 

tion  of  the  apsis,  the  eccentricity,  and  the  periodic  time. 

Ex.  6.  A  body  is  projected  at  a  given  distance  from  the 
center  of  force  with  a  given  velocity,  and  in  a  direction  per- 
pendicular to  the  distance,  when  the  force  is  repulsive  and 
varies  inversely  as  the  cube  of  the  distance.  Find  the  path  of 
the  body. 


HYDROSTATICS. 


CHAPTER   I. 

377.  Matter  exists  in  the  various  states  of  solid,  fluid,  and 
aeriform.  In  solid  bodies  the  homogeneous  integrant  particles 
cohere  firmly,  and  do  not  admit  of  interchange  of  position. 
The  homogeneous  integrant  constituents  of  fluids  possess  less 
cohesion,  and  change  their  position  among  each  other  on  the 
application  of  a  moderate  force. 

Fluids  differ  from  each  other  in  the  degree  of  cohesion  of 
their  constituent  parts  and  the  facility  with  which  they  will 
yield  to  the  action  of  an  impressed  force. 

378.  A  perfect  fluid  is  one  in  which  the  cohesion  of  the  con- 
stituent particles  is  so  feeble  that  the  least  force  is  capable  of 
effecting  a  separation  and  causing  motion  among  each  other 
in  all  directions.  Perfect  fluids  are  the  only  ones  which  will 
be  made  the  subject  of  investigation. 

Fluids  are  divided  into  two  classes,  incompressible  or  liquid, 
and  compressible  or  aeriform. 

379.  Incompressible  fluids  are  those  which  retain  the  same 
volume  under  a  variable  pressure.  They  may  be  made  to  take 
an  infinite  variety  in  form,  which  form  they  will  retain  when 
acted  upon  by  no  external  pressure  nor  accelerating  force. 
Of  this  class  water  may  be  regarded  as  the  type. 

380.  Compressible  fluids,  of  which  atmospheric  air  is  the 
type,  change  at  once  in  form  or  volume  with  every  variation 
of  pressure  which  they  sustain,  and  return  again  to  the  same 
form  and  volume  when  the  same  circumstances  of  pressure  and 
temperature  are  restored. 


EQUALITY     OF     PRESSURE. 


237 


PRINCIPLE   OF    EQUALITY   OF   PRESSURE. 

381.  Prop.  A  force  or  pressure  applied  to  a  given  portion  of 
the  surface  of  a  fluid  is  transmitted,  undiminished  in  intensity, 
to  every  other  equal  portion  of  the  surface. 

A  force  impressed  on  a  solid  is  effective  only  in  the  direc- 
tion of  its  action,  and  is  sustained  by  an  equal  force  impressed 
in  an  opposite  direction.  Applied  to  a  fluid  the  force  is  effect- 
ive in  every  direction,  and  can  be  sustained  only  by  forces  ap- 
plied to  every  point  in  the  surface  of  the  fluid. 

Let  the  annexed  figure 
represent  a  section  of  a 
vessel  filled  with  a  fluid 
without  weight.  In  this 
case,  supposing  the  press- 
ure of  the  atmosphere  re- 
moved also,  the  surface  of 
the  fluid  would  be  subject 
to  no  pressure,  and  would 
retain  its  form  if  the  sides  of 
the  vessel  were  removed. 

Let  an  aperture  of  a  giv- 
en size,  as  one  square  inch, 
be  made  in  either  side,  AB, 
and  a  piston  be  accurately  fitted  to  it,  and  supposed  to  move 
without  friction.  Now,  since  the  particles  are  without  friction, 
a  force  P,  of  one  pound,  applied  to  this  piston,  will  act  upon 
the  stratum  of  particles  in  contact  with  its  base,  and  this  stra- 
tum upon  the  next,  and  so  on.  The  force  will  therefore  be 
transmitted  to  the  opposite  side,  EF,  and  will  cause  motion  in 
the  equal  piston  P'l5  unless  counterpoised  by  an  opposite  force 
of  one  pound.  If  this  counterpoise  be  applied,  then,  since  the 
particles  move  freely  among  each  other  without  resistance 
from  friction  or  cohesion,  the  particles  in  the  direction  of  the 
applied  forces  will  act  on  those  lying  without,  and  will  commu- 
nicate motion  to  them,  unless  resisted.  A  motion  will  therefore 
be  communicated  to  any  equal  piston  T'[,  unless  sustained  by  a 


238  HYDROSTATICS. 

pressure  of  one  pound.  Hence  both  pistons  P',  and.  P','  will 
sustain  the  same  pressure  by  the  action  of  P,. 

Further,  if  the  pistons  P',  and  P"  were  contiguous,  or  one 
of  them  twice  the  size  of  P,,  then  it  is  obvious  that  a  force  of 
two  pounds  would  be  necessary  to  preserve  the  equilibrium  ; 
or,  if  the  base  of  the  piston  P4  be  four  times  that  of  Pj,  then  a 
force  of  four  pounds  would  be  necessary  to  keep  P4  at  rest; 
and  if  the  base  of  one  piston  be  n  times  that  of  the  other,  the 
pressure  on  the  former  will  be  n  times  that  on  the  latter,  or  a 
force  of  one  pound  will  produce  a  pressure  of  n  pounds,  where 
n  may  be  the  ratio  less  one,  of  the  whole  surface  of  the  vessel 
to  the  base  of  the  piston. 

Cor.  1.  The  pressures  Yx  and  P„  on  any  two  portions  A, 
and  A„  of  the  surface  will  be  proportional  to  their  areas,  or 

£=£=«.  (134) 

Also,  the  normal  pressure^  on  a  unit  of  surface  will  be 
P       P 

Cor.  2.  Every  stratum  of  particles  in  the  interior,  of  the 
same  dimensions  as  the  base  of  the  piston,  wherever  situated 
and  however  inclined,  is  subject  to  a  pressure  equal  to  that  ap- 
plied to  the  piston  ;  and  since  the  fluid  is  without  weight,  every 
particle  presses  and  is  pressed  equally  in  every  direction. 

Cor.  3.  If  all  the  pistons  except  P  and  P'  be  firmly  fixed, 
and  P  be  forced  in,  since  the  fluid  is  incompressible,  whatever 
fluid  is  displaced  by  P  will  be  forced  into  the  tube  P'.  If  h  and 
h'  be  the  spaces  through  which  the  pistons  move,  a  and  a'  the 
areas  of  their  bases,  then 

ha=h'a'. 

But  P  and  P'  being  the  equilibrating  pressures  on  these  areas, 

«_P 

a'~P'; 
hence  hP=h'¥',  or  AP-/i'P'=0, 

or  the  principle  of  virtual  velocities  applies  to  fluids  in  equilib- 
rium. 


SURFACES     OF     EQUILIBRIUM.  239 


SURFACES   OF    EQUILIBRIUM. 

382.  Prop.  The  free  surface  of  a  fluid  subjected  to  the  action 
of  an  accelerating  force  is,  when  in  equilibrium,  perpendicular 
to  the  direction  of  the  force. 

A  liquid  void  of  gravity  will  take  and  retain  any  form  im 
pressed  upon  it ;  but  when  subject  to  the  action  of  an  acceler- 
ating force,  a  containing  vessel  will  be  necessary  in  order  to 
preserve  a  coherent  mass. 

Let  ABCD  be  the  section  of  a  ves- 
sel containing  a  fluid  subject  to  the  ac- 
tion of  gravity,  the  base  DC  being  hori- 
zontal. If  any  portion  of  the  free  sur- 
face, as  abc,  have  a  direction  not  per- 
pendicular to  the  direction  of  gravity, 
this  force  may  be  resolved  into  two 
components,  one  of  which  is  parallel  to  this  surface  ;  and  since 
the  particles  are  free  to  move  in  that  direction  as  down  an  in- 
clined plane,  they  will  yield  to  this  component.  But  whenever 
every  portion  of  the  surface  is  perpendicular  to  the  direction 
of  gravity  this  force  will  have  no  component  in  the  direction 
of  the  surface,  and  every  portion  of  it  will  be  urged  vertically 
downward  with  equal  intensity. 

Cor.  Hence  the  surfaces  of  fluids  at  rest,  and  acted  on  only 
by  gravity,  are  horizontal.  But  since  the  directions  of  gravity, 
acting  on  particles  remote  from  each  other,  are  convergent  to 
the  center  of  the  earth  nearly,  the  surfaces  of  large  masses  of 
fluid  are  not  plane,  but  curved,  and  conform  to  the  general 
figure  of  the  earth. 

383.  Prop.  If  a  vessel  containing  a  fluid  be  moved  horizontal- 
ly with  a  constant  accelerating  force,  the  surface  will  take  the 
position  of  an  inclined  plane. 

Let  the  vessel  ABCD  containing  a  fluid  be  moved  horizon- 
tally with  a  uniformly  accelerating  force  P.  Then  any  ele- 
ment, k,  of  the  surface,  whose  weight  is  w,  and  mass  -  (22). 

O 

will  be  urged  by  its  weight  iv  in  the  direction  kg,  and  by  it? 


240 


HYDROSTATICS. 


inertia  P—  in  the  direction  kk,  and 

g 
therefore   by  their  resultant  in  the 

direction  ko.     As  the  same  is  true 
of  every  element  of  the  surface,  the 
whole  surface  (Art.  382)  will  be  per- 
pendicular to  ko. 
Let  the  ansrle  of  inclination  to  the  horizon  HIM=I.     Then 


since  HIM= okg, 


tan.  o/co-=tan.  1= 


P» 


w 


384.  Prop.  If  a  vessel  containing  a  fluid  be  made  to  revolve 
uniformly  about  a  vertical  axis,  the  surface  of  the  fluid  ivill  take 
the  form  of  a  paraboloid  of  revolution. 

Let  the  vessel  ABCD  revolve  uni 
formly  about  the  vertical  axis  XX  j. 
Any  element  k  of  the  surface  will  be 
urged  horizontally  by  a  centrifugal  force 
directed  from  the  axis,  and  vertically 
downward  by  gravity.  Let  w  be  the 
angular  velocity  of  the  fluid,  w  the  weight 
Xi  of  the  element  k,  and  y  its  distance  AM 

from  the  axis.     Then  (82)  we  have,  for  the  centrifugal  force  P 
of  this  element, 

g 
Being  urged  vertically  downward  by  its  weight  w,  the  surface 
of  the  element  will  be  perpendicular  to  the  resultant  AR=R 
of  these  two  forces.     If  this  resultant  be  produced  to  meet  the 
axis  in  N,  we  have 


and 


or 


LRkg=Lk'NM, 
Rg     AM 


kg 

0)  v— 
g. 
w 


'MN' 

y 

"MN" 


PRESSURES     ON     IMMERSED     SURFACES. 


241 


.'.  the  subnormal  MN=— =  a  constant,  a  property  of  the 
parabola.     Hence  the  surface  is  a  paraboloid  of  revolution. 


NORMAL    PRESSURES    ON    IMMERSED    SURFACES. 

385.  Prop.  The  pressure  on  the  horizontal  base  of  a  vessel 
containing  an  incompressible  fluid  is  proportional  to  its  depth 
below  the  surface  of  the  fluid,  and  independent  of  the  form  of  the 
vessel. 

Let  HO  be  the  free  surface  of  the  fluid  in 
the  vessel  ABCD,  whose  base  DC  is  horizon- 
tal, and  suppose  the  fluid  divided  into  hori- 
zontal strata  of  small  but  equal  thickness. 

Let  «,,  a2,a3,  &c,  denote  the  successive 
strata, 

A0,  A,,  A2,  the  units  of  surface  in  each, 

P0,  P,,  P2,  the  whole  pressure  on  each, 

p  0,  p  x,  p2,  the  pressure  on  a  unit  of  each, 
a  the  common  thickness  of  the  strata,  and  p  the  density  of  the 
fluid.     Since  the  thickness  of  each  stratum  is  supposed  indef- 
initely small,  the  upper  and  lower  surfaces  of  each  may  be  re- 
garded as  equal. 

Now  the  weight  of  al  is  wx—gpaAQ  (24), 
"      "         "        "  a2  is  w2=gpaAlf 

"      "         "        "  an  is  wn  =gpaAn_t. 
The  pressure  on  al  is  P0, 

P 

on  a  unit  of  al  is  -r^—p0  (135), 

on  a2  is  P,=P0+^1=P0+5,paA0, 

.      .      .P.      P0    gpaA. 

on  a  unit  oi  an  is  -r- =-r--\ — r or  p  1  =p 0  +gpa, 

A,      A,         Aj 

on  a 3  is  Pa=P1+w2=P1  +gpaAlf 

.    P2     P,  ,gpaA,    * 
on  a  unit  of  a 3  is  t-=t — 1 — t —  orPa  ~P  i  +gp-a=p9+gp.2a. 


on  a„+1  is  P(,=P,_I+u;Il=Pfl_1-f-g,paAl_„ 

Q 


242  HYDROSTATICS. 

.      r         .    P„     P,,-!    gpaAn_! 
on  a  unit  01  rt„+1  is  -7- =-7 — ' a or  p,=p0+gp.?ia. 

But  na—h  is  the  depth  of  a,l+l  below  the  surface  of  the  fluid, 
and  if  the  upper  surface  of  a„+l  represent  the  surface  of  the 
base,  A  the  number  of  units  in  the  base,  then  the  pressure  p  on 
a  unit  of  the  base  will  be 

P=P0+gph> 
and  the  whole  pressure  on  the  base 

V=Ap=Ap0+gphA=VQ+gPhAorh.      (136) 

Cor.  1.  Since  h A  is  equal  to  a  prism  whose  base  is  A  and 
height  h,  and  gp/iA  is  its  weight,  if  we  disregard  P0,  which 
may  represent  the  pressure  of  the  atmosphere  on  the  surface 
of  the  fluid,  we  have,  for  the  pressure  of  the  fluid  on  the  base 
of  the  vessel,  the  weight  of  a  column  of  the  fluid  ivhose  base  is 
that  of  the  vessel,  and  height  the  height  of  the  surface  of  the  fluid 
above  the  base.  It  is  obviously  immaterial  whether  the  surface 
pressed  is  that  of  the  base  of  the  vessel  or  a  horizontal  surface 
of  an  immersed  solid. 

Cor.  2.  Since  a  cubic  foot  of  water  weighs  1000  oz.  =  62.5 
lbs.,  we  have,  for  the  pressure  on  the  base  of  any  vessel  con- 
taining water, 

P=62.5  AA  lbs.,  (137) 

where  h  is  the  height  in  feet  of  the  surface  of  the  water  above 
the  base,  and  A  the  number  of  square  feet  in  the  base. 

Cor.  3.  The  pressure  on  every  portion  of  a  horizontal  stra 
turn  of  the  fluid  will  be  the  same,  and  since  this  pressure  is 
transmitted  equally  in  every  direction,  the  pressure  on  every 
element  of  the  sides  of  the  vessel  having  the  same  depth  will 
be  equal  to  that  on  the  surface  of  the  stratum.  If  the  sides  of 
the  vessel  are  inclined,  this  pressure  will  be  the  normal  press 
ure  on  the  sides  at  that  depth.  Let  the  annexed  figures,  A, 
B,  C,  D,  represent  vessels  of  different  forms  and  capacities 
with  equal  bases.  Then  the  pressures  on  the  base  of  each 
will  be  equal  when  filled  to  a  common  height,  and  the  points 
a,  b,  c,  e,  g,  h,  k,  I,  m,  having  a  common  depth,  will  be  equally 
pressed  normally  to  the  surfaces.  In  B  the  horizontal  surface 
df  of  the  vessel  will  experience  the  same  pressure  vertically 


PKESSUKES     ON      IMMERSED     SURFACES.  243 

A  W  C  D 


upward  as  that  to  which  the  stratum  fg  of  the  fluid  is  subject. 
Moreover,  the  stratum  fg  being  in  equilibrium,  must  be  press- 
ed equally  upward  and  downward.  The  downward  pressure 
is  equal  to  the  area  of  the  section  fg  into  the  depth  h  of  the 
section  below  the  surface  of  the  fluid  in  the  vertical  branch. 
But  the  upward  pressure  is  due  to  the  pressure  of  the  fluid  in 
the  inclined  branch  ;  and  since  the  area  of  the  section  fg  is 
common  to  the  measures  of  both  pressures,  the  vertical  heights 
of  the  fluid  in  both  branches  are  necessarily  equal. 

386.  Prop.  The  normal  pressure  on  any  plane  surface  in- 
clined to  the  horizon  is  proportional  to  the  depth  of  its  center 
of  gravity  below  the  surface  of  the  fluid. 

Let  k  be  any  indefinitely  small  element  of  the  immersed  sui 
face,  x  its  depth,  and  p  the  pressure  it  sustains.     By  the  pre- 
ceding proposition, 

p—gpxk; 

and  since  the  pressure  is  the  same  in  every  direction,  p  will  be 
the  pressure  normal  to  this  element,  whatever  be  its  position 
or  inclination.  The  expression  for  the  normal  pressure  on 
every  other  element  will  be  of  the  same  form.  Hence  the 
whole  normal  pressure  P  will  be 

P= I,gpxk=gplxk. 

Let  x  be  the  depth  of  the  center  of  gravity  of  the  immersed 
surface  below  the  surface  of  the  fluid,  and  A  its  area.  Then 
(29),  Ax=lxk; 

...  Y=gpAx<xx.  (138) 

Cor.  Since  gpAx  is  the  weight  of  a  prism  of  the  fluid  whose 
base  is  A  and  altitude  x,  the  normal  pressure  on  any  immersed 
surface  inclined  to  the  horizon,  or  on  any  side  of  the  contain- 
ing-vessel,  is  equal  to  the  weight  of  a  fluid  prism  whose  base 


ti44 


HYDROSTATICS. 


is  the  surface  pressed,  and  height  the  depth  of  its  center  oi 
gravity  below  the  fluid  surface,  or,  if  the  fluid  be  water, 
P=62.5  Ailbs.  (139) 

387.  Prop.  To  find  the  pressure  on  any  immersed  surface  or 
side  of  a  vessel  in  a  given  direction. 

Let  P  be  the  normal  pressure  on 
the  inclined  side,  CDFE  =  A  of  the 
vessel  CBE,  and  let  it  be  required 
to  determine  the  horizontal  and 
vertical  pressures  on  A.  Let  a  be 
the  angle  which  the  side  A  makes 
with  a  vertical  plane,  and  (3  its  in- 
clination  to   the   horizon.      Then, 

since  PAP,=a  and  PAP2=/3,  resolving  P  horizontally  and 

vertically,  we  have 

P,=P  cos.  a—gpxA  cos.  a, 

and  P2  =P  cos.  P=gpxA  cos.  P. 

But  CDHG=A,  is  thn  projection  of  the  side  A  on  a  vertical 

plane,  and  EFHG= A2  is  the  projection  of  the  same  side  on  a 

horizontal  plane. 

Therefore,     A,=A  cos.  a  and  A2=A  cos.  /3, 
and  F  l—gpAlx    and  Y2—gpK2x. 

Hence  the  horizontal  pressure  on  A  is  equal  to  the  weight 
of  a  prismatic  volume  of  the  fluid  whose  base  is  the  vertical 
projection  of  A  and  height  the  depth  of  its  center  of  gravity ; 
and  vertical  pressure,  a  volume  whose  base  is  the  horizontal 
projection,  and  height  the  depth  of  the  center  of  gravity  of  A. 
And  generally  the  pressure  on  any  immersed  surface  in  a  given 
direction  is  equal  to  the  weight  of  a  prismatic  column  of  the 
fluid  whose  base  is  the  projection  of  the  surface  on  a  plane  per- 
pendicular to  the  given  direction,  and  height  the  depth  of  the 
center  of  gravity  of  the  surface  below  the  surface  of  the  fluid. 

388.  Prop.  To  find  the  resultant  pressure  of  a  fluid  on  the  in- 
terior surface  of  the  containing  vessel. 

Let  ABCD  be  a  section  of  a  vessel  of  any  form  filled  with 
a  fluid.     When  the  sides  of  the  vessel  are  curved,  the  normal 


PRESSURES     ON     IMMERSED     SURFACES. 


215 


pressures  on  the  elements  of  the 
surface  will  not  be  parallel.  But 
we  may  resolve  the  normal  press- 
ure on  each  element  into  vertical 
and  horizontal  components,  and 
each  horizontal  component  into 
two  others  parallel  to  two  rectan- 
gular axes.  We  shall  then  have 
to  find  the  resultants  of  three  sets 
of  parallel  forces,  and,  finally,  the  resultant  of  these  three  re- 
sultants. We  may,  however,  determine  the  resultant  more 
simply,  as  follows  : 

1°.  The  resultant  of  the  horizontal  pressures. 

Let  HK  be  the  intersection  of  a  vertical  plane  by  the  plane 
ABCD  produced,  and  let  k  be  the  vertical  projection  of  any 
element  kr  on  this  plane.  Now  the  normal  pressure  on  the 
element  kx  of  depth  a1k1=h  is  hkx,  and  the  horizontal  press- 
ure {Art.  387)  is  hk.  The  horizontal  pressure  on  each  ele- 
ment being  of  the  same  form,  the  whole  horizontal  pressure 
perpendicular  to  HK  of  the  surface  convex  toward  HK  will 
be  Z.hk.  But  to  every  element  &,  of  the  surface  convex  to- 
ward HK  there  is  a  corresponding  element  k2  of  the  surface 
concave  toward  HK,  the  projection  k  of  which  is  the  same. 
Hence  the  horizontal  pressure  on  k2  is  —  hk,  and  the  pressure 
on  the  whole  surface  concave  toward  HK  is  —Zhk.  The  re 
sultant,  therefore,  of  all  the  horizontal  pressures  perpendicular 
to  KH  is  2,.hk—2.hk=0.  Since  the  same  is  true  on  whatever 
side  of  the  vessel  the  plane  HK  be  drawn,  the  resultant  of  all 
the  horizontal  pressures  will  be  zero. 

2°.  The  resultant  of  the  vertical  pressures. 

The  pressure  on  the  element  kl  vertically  downward  is  the 
weight  of  a  filament  of  the  fluid  whose  base  is  kx  and  height 
a1kl.  The  pressure  on  k'  vertically  upward  is  equal  to  the 
weight  of  a  filament  of  the  fluid  whose  base  is  k'  and  height 
ajt1.  The  resultant  of  the  pressures  on  the  corresponding  ele- 
ments k'  and  kx  is  therefore  equal  to  the  weight  of  the  fila- 
ment of  fluid  k'kx.  In  the  same  manner,  the  resultant  of  the 
vertical  pressures  on  any  two  corresponding  elements  will  be 


246 


HYDROSTATICS. 


the  weight  of  the  filament  of  fluid  whose  bases  are  these  ele« 
ments.  The  resultant  of  all  the  vertical  pressures  will  then 
obviously  be  the  sum  of  the  weights  of  these  filaments,  or  the 
weight  of  the  fluid. 

Hence  the  resultant  of  all  the  partial  pressures  on  the  in- 
terior surface  of  the  vessel  is  equal  to  the  weight  of  the  fluid, 
directed  vertically  downward,  and  will  pass  through  the  cen- 
ter of  all  the  parallel  vertical  pressures  {Art.  44),  or  the  center 
of  gravity  of  the  fluid  {Art.  94). 

Cor.  Since  the  horizontal  pressures  balance,  there  will  be 
no  resultant  pressure  to  cause  motion  in 
the  vessel  horizontally.  If,  however, 
an  aperture  be  made  on  one  side  of  the 
vessel,  the  pressure  on  this  portion  of 
the  vessel  will  be  removed,  and  the 
pressure  on  the  corresponding  opposite 
portion  will  tend  to  produce  motion  in 

that  direction,  and  if  the  vessel  be  free  to  move,  motion  will 

actually  ensue. 

389.  Prop.  To  find  the  resultant  pressure  on  any  solid  im- 
mersed in  a  fluid,  and  its  point  of  application. 

1°.  The  horizontal  pressure  on  any  element  of  the  immersed 

solid  is  the  same  as  that  on  the 
vertical  projection  of  the  element 
{Art.  387),  and  the  entire  press- 
ure on  each  of  the  opposite  sur- 
faces of  the  solid  will  be  equal  to 
that  on  their  vertical  projections. 
But  the  projections  of  any  two  op- 
posite surfaces  on  the  same  vertical  plane  are  equal.  Hence 
the  horizontal  pressures  on  any  two  opposite  surfaces  are  equal, 
and  the  resultant  of  the  horizontal  components  of  the  pressures 
in  every  direction  is  zero. 

2°.  The  vertical  downward  pressure  p'  on  any  element  k' 
of  the  upper  surface  is  equal  to  the  weight  of  a  filament  of  the 
fluid  ah'  whose  base  is  k'  and  altitude  ak'=h',  or  if  It  be  the 
horizontal  transverse  section  of  the  filament  through  the  cen 


PRESSURES     ON     IMMERSED     SURFACES.  241 

ter  of  gravity  of  k',  p'—gph'k.  But  the  sum  of  all  the  filaments 
resting  on  the  upper  surface  of  the  solid  is  equal  to  the  volume 
V  of  the  fluid  vertically  above  the  solid,  and  the  whole  down- 
ward pressure  P'  is  the  weight  of  this  volume,  or 
V'=Z.p>=gpZ.h'k=gpV. 
In  the  same  manner,  the  vertical  upward  pressure  p  x  on  the, 
corresponding  element  kx  of  the  lower  surface  is  equal  to  the 
weight  of  a  filament  akl  of  fluid  whose  base  is  kx  and  height 
ak^—J^,  or pi—gph1k.  But  the  sum  of  all  these  filaments  is 
the  volume  V,  of  the  solid  and  fluid  above  the  upper  surface, 
and  the  whole  upward  pressure  P,  is  the  weight  of  this  volume 
of  fluid,  or 

Now  h1—h'=h  is  the  length  of  the  filament  k'k^  of  the  solid 
p}-ip'=p  its  weight,  and  V l—V'=Y  is  the  volume  of  the 
solid. 

Therefore,  the  difference  P  of  the  upward  and  downward 
pressure  is 

P=P1-P/=2.p1-2.p'=:£.;?, 
and  'P=gp2.h1k—gp2.h'k=gp2hk=gpV;  * 

...  Y=gpV=2p,  (140) 

or  the  resultant  pressure  is  vertically  upward,  and  equal  to  the 
weight  of  a  mass  of  the  fluid  of  the  same  volume  as  the  solid. 
3°.  To  find  the  point  of  application  of  this  resultant.  Let 
the  distance  of  the  point  of  application  of  p=hk  from  the  ver- 
tical plane  HK  be  x,  and  the  distance  of  the  point  of  applica- 
tion of  the  resultant  P  be  x.  Then,  since  the  moment  of  the 
resultant  equals  the  sum  of  the  moments  of  the  components, 
l*x=2.px, 

Z.px     S.px 
and  *=-£-=~^f  <141) 

But  (29)  the  point  thus  determined  is  the  center  of  gravity 
of  the  displaced  fluid. 

Hence  the  resultant  of  all  the  pressures  on  the  immersed 
solid  is  equal  to  the  weight  of  the  displaced  fluid,  acts  vertical- 
ly upward,  and  its  point  of  application  is  the  center  of  gravity 
of  the  displaced  fluid. 


248  HYDROSTATICS. 

390.  Prop.   To  find  the  conditions  of  equilibrium  rf  an  xm 
mersed  solid. 

The  conditions  of  equilibrium  involve  a  consideration  of  the 
weight  of  the  body.  Let  V  be  its  volume,  and  a  its  density. 
Its  weight  will  be  goV.  The  body  will  be  urged  downward 
by  a  force  equal  to  goV  applied  at  its  center  of  gravity,  and 
by  Art.  389  it  is  urged  upward  by  a  force  equal  to  gpV,  since 
the  volumes  of  the  solid  and  displaced  fluid  are  equal. 

In  order  to  equilibrium,  then,  1°.  These  forces  must  be  equal; 
and,  2°.  Their  lines  of  direction  coincident. 

The  first  condition  gives 

gaV-gpV> 

or  o=p ; 

that  is,  their  densities  must  be  the  same. 

Cor.  1.  If  o  be  not  equal  to  p,  the  body  will  ascend  oi;  de- 
scend by  a  force  equal  to  g(p—o)V,  according  as  p—o  is  posi- 
tive or  negative.  If  p>a,  the  body  will  rise  to  the  surface,  and 
be  but  partially  submerged.  Let  Vj  be  the  displaced  fluid,  or 
the  part  of  the  solid  immersed  when  the  equilibrium  is  restored 
Then 

gpV,=g°V,  (142) 

and  V  :  V ,  =p  :  a  ; 

or  the  whole  volume  of  the  solid  is  to  the  part  immersed  as  the 
density  of  the  fluid  is  to  the  density  of  the  solid. 

Cor.  2.  If  the  centers  of  gravity  of  the  solid  and  displaced 
fluid  be  not  in  the  same  vertical  line,  the  body  will  be  acted 
upon  by  two  parallel  forces  in  opposite  directions,  and  will 
cause  the  body  to  turn  round.  The  point  of  application  of  the 
resultant  of  these  forces  may  be  found  by  Art.  29. 

391.  Def.  The  section  of  a  floating  body  made  by  a  plane 
coincident  with  the  surface  of  the  fluid  is  called  the  pla?ie  of 
flotation.  The  line  joining  the  centers  of  gravity  of  the  solid 
and  of  the  displaced  fluid  is  called  the  axis  q  'flotation. 

DEPTH    OF   FLOTATION. 

392.  Prop.  To  find  the  depth  of  flotation  when  the  volum* 
and  density  of  the  body  are  known. 


DEPTH     OF     FLOTATION.  249 

Let  V  be  the  volume  of  the  body,  o  its  density,  Vj  the  volume 
of  the  displaced  fluid,  and  p  the  density  of  the  fluid.    Then  (142) 
goY=gpV„ 

and  V.=-V. 

p 

Now  whenever  V,  can  be  determined  in  terms  of  the  depth 
of  flotation  x,  this  expression  will  suffice  to  determine  x. 

If  the  solid  be  a  right  cylinder  whose  axis  a  is  vertical,  and 

the  radius  of  whose  base  is  r,  we  have 

Y  l=irr'x, 

and  Y—nr'a; 

o 
.'.  x—-a. 
P 

If  the  body  be  a  right  cone  with  the  axis  vertical  and  vertex 

downward,  let  r  be  the  radius  of  the  base,  and  a  the  altitude 

T 

Then  the  radius  of  the  plane  of  flotation  is  -x.     Hence 

a 

r2 

V  =l7T— X3 

V  1  3  Q*  ' 

and  Y=^Trr>a; 

3  /a 

.'.  x=a\/  -. 

V   p 

If  the  vertex  of  the  cone  be  upward, 

3/      a 
y  x—a\     I— . 

cn  v      p 

393.  Prop.  To  find  the  positions  of  equilibrium  of  a  right 
triangular  prism  when  the  axis  is  horizontal. 

Whenever  a  body  can  be  conceived  to  be  generated  by  the 
motion  of  a  plane  surface  perpendicular  to  itself,  and  the  body 
floats  with  its  axis  horizontal,  it  is  evident  that,  in  whatever 
position  it  be  turned,  its  center  of  gravity  and  that  of  the  part 
immersed  will  lie  in  the  same  vertical  section ;  and,  further, 
that  the  center  of  gravity  of  the  body  will  be  at  the  center  of 
gravity  of  the  section,  and  the  center  of  gravity  of  the  im- 
mersed part  at  the  center  of  gravity  of  its  section.  Also,  the 
ratio  of  the  part  immersed  to  the  whole  mass  is  the  same  as 


250 


HYDROSTATICS. 


the  ratio  of  their  sections.     We  can  therefore  limit  the  investi 
gation  to  the  positions  of  equilibrium  of  one  of  the  generating 
sections. 

A  Let  ABC,  Fig.  1,  be  a 

generating  section  when 
only  one  angle  is  im- 
mersed, and  Fig.  2  when 
two  angles  are  immersed. 
LetAB=c,AC=6,BC=a, 
AQ,=;r,AP=?/,andlet<7be 
the  ratio  of  the  density  of 
the  solid  to  that  of  the  fluid. 
Now  from  (142)  we  have,  in  Fig.  1, 

AQP     \xy  sin.  A_xy 


ABC 


\bc  sin.  A 
xy=cbc. 


be' 


(a) 


In  Fig.  2, 


BQPC    ABC-AQP    bc-xy 


ABC 


be 


Q>) 


ABC 
.'.  xy=bc(\—  a). 

Since  (b)  is  derived  directly  from  (a),  by  changing  a  into 
1  —  a,  we  may  examine  the  case  of  Fig.  1,  and  deduce  that  of 
Fig.  2  from  the  result,  by  changing  a  into  1  — a. 

Bisect  BC  in  M,  and  QP  in  m.  Join  AM  and  Km.  Take 
MM'=plA  and  mm'=\mK.  M'  and  m'  are  the  centers  of 
gravity  of  ABC  and  AQP.  Join  Mm  and  M'm'.  It  is  obvious 
that  the  center  of  gravity  of  BQPC  is  in  the  line  M'm'  produced. 
Now  the  second  condition  of  equilibrium  requires  that  M'm' 
should  be  vertical,  and  since  Mm  is  parallel  to  M'm',  Mm  is 
perpendicular  to  QP  and  MQ=MP.  Let  AM=A,  MAQ=0, 
and  MAP=0.     Then 

MQ?=x"-+h2-2hx  cos.  6, 
and  MF=ys-fA*-2Aycos.^. 

Hence  x*  — y1  —  2hxcos.d+2hy  cos.  0=0.  (c) 

Substituting  in  (c)  the  value  of  y  from  (a),  and  reducing, 

xi-2h  cos.  d.xz+2abch  cos.  0.ar— ff'&V=0,  (d) 


DEPTH     OF     FLOTATION.  251 

and  changing  a  into  1  —  c,  we  have  for  Fig.  2, 

xi-2h  cos.  6.x2 +2(1  -o)bch  cos.  <£.a;-(l-a)W=0.  (e) 

The  values  of  a;  deduced  from  (rf)  and  (e),  and  the  correspond 
ing  values  of  y  from  (a)  and  (b),  will  give  the  positions  of  equi 
librium  required. 

Now,  since  the  degree  of  the  equation  is  even  and  the  absol- 
ute term  is  negative,  there  are  at  least  two  possible  roots,  one 
positive  and  the  other  negative.  The  other  two  roots  may  be 
real  or  imaginary.  If  real,  Descartes'  rule  of  signs  indicates 
that  three  will  be  positive  and  one  negative.  Hence,  since  the 
values  of  x  and  y,  which  are  applicable  to  the  question,  are 
necessarily  positive,  equations  (d)  and  (e)  indicate  no  more 
than  three  positions  of  equilibrium.  The  values  of  a:  must  also 
be  less  than  b,  and  when  subtituted  in  (a)  and  (b)  give  for  y 
values  less  than  r. 

Let  us  take  the  case  of  an  equilateral  triangle. 

Then    6=<p— 30,  a=b=c,  and  h=b  cos.  6=c  cos.  0. 

These  values,  substituted  in  (d),  give 

x4  -  lax'  +  \aa%x  -  a2 a* = 0, 
or  (x2— oa?)  {x2  — lax + ca2)— {).  (J) 

This  equation  is  satisfied  by  putting 
x2  —  oa2=0; 
whence  x  =a^o, 

the  negative  value  of  £  being  inapplicable. 

Since  #<1,  we  have  x<a,  and  the  value  of  x  indicates  an 
actual  position  of  equilibrium.     But  from  (a)  we  have 

or.  a2      o.a2  — 

y= = — —=a  vo—x, 

x       aVo 

or  one  side  of  the  triangle  is  parallel  to  the  surface  of  the  fluid. 

Equation  (/)  is  also  satisfied  by  making 

x2  —  lax+oai=0 ; 


a 


whence  x  =-{3±  V9—  16a}.  (g) 

Now,  in  order  that  these  values  of  a;  may  be  real,  we  must 
have 

16o<9  or  16cr=9. 


252  HYDROSTATICS. 

9  9 

Hence  a<—  or  °=Jq, 

9 

or  o  can  not  be  greater  than  — . 

But  if  V9—  16<t>1,  then  x>a,  which  is  inconsistent  with  tl# 
supposition  that  only  one  angle  is  immersed.  The  greatest 
value,  therefore,  which  V9—  IGo  can  have  is  unity,  and  in  this 
case  the  second  angle  will  lie  in  the  surface  of  the  fluid. 

Putting  V9— 16a=l, 

1 
we  get  <r=-, 

for  the  least  value  a  can  have  when  one  angle  only  is  im- 
mersed.    The  limits  of  a  for  this  case  are,  therefore, 

1  A     9 

-  and  — . 

2  16 

If  in  (g)  we  cnange  o  into  1  —  o,  we  get 


z= -\3±V  9-16(l-o)] 


=-{3±Vl6(T-7}.  (A) 

From  which  it  may  be  shown  that  the  limits  of  the  value  ot 
u  for  this  case  are 

—  and  -. 
16  2 

Since  equations  (d)  and  (e)  may  have  three  real  positive 
roots  each,  there  may  be  three  positions  of  equilibrium  for 
every  single  angle  immersed,  and  three  for  every  two  angles 
immersed,  and  therefore  eighteen  in  the  whole.  But  in  the 
particular  form  considered  above  these  are  not  all  possible, 

except  a=-,  in  which  case  x=a  or  y=a;  that  is,  either  the 

angle  B  or  C  lies  in  the  surface  of  the  fluid.  This  would  ren- 
der six  of  the  positions  pertaining  to  Fig.  1  the  same  as  six  of 
those  pertaining  to  Fig.  2,  making  but  twelve  really  different 

1  9 

ones.     If  tf>„  and  <7<t^>  there  will  be  nine   positions  with 


? 


tyyf 


ff 


f 


.  f 


CENTER     OF     PRESSURE. 


253 


one  angle  immersed  and  three  with  two  immersed ;  if  ct<- 

7  7 

and  o>—,  nine  of  the  latter  and  three  of  the  former;  if  a<— 

lo  lb 

9 

or  0>— ,  there  will  be  only  three  of  each. 
lo 


D 


A 


CENTER    OF    PRESSURE. 

394.  Def.  The  center  of  pressure  in  any  immersed  surface 
is  the  point  of  application  of  the  resultant  of  all  the  pressures 
upon  it.  It  is  therefore  that  point  in  an  immersed  surface  or 
side  of  a  vessel  containing  a  fluid,  to  which,  if  a  force  equal 
and  opposite  to  the  resultant  of  all  the  pressures  upon  it  be 
applied,  this  force  would  keep  the  surface  at  rest. 

395.  Prop.  To  find  the  center  of  pressure  on  any  immersed 
plane  surface. 

Let  the  immersed  surface 
be  the  inclined  side  AE  of 
the  vessel  ABE  supposed  to 
be  filled  with  a  fluid.  Let  x 
and  y  be  the  co-ordinates  of 
any  element  k  referred  to 
the  rectangular  axes  AY  and 
AX,  and  hk=h  the  vertical 
distance  of  the  element  be- 
low the  surface  of  the  fluid. 
Then  the  normal  pressure  jo  on  k  will  be 

p=gphk. 

The  pressure  on  each  element  will  be  of  the  same  form,  and 
their  sum  or  resultant  R  will  be 

R=2./>. 

The  moment  of  each  partial  pressure,  in  reference  to  the 
plane  through  AY  perpendicular  to  AX  (Art.  46),  will  be  px, 
and  the  moment  of  each,  in  reference  to  the  plane  through  AX 
perpendicular  to  AY,  will  be  py.     Hence,  if  x  and  y  are  the 


iy 

f> 

7/l/X;.- ii^_ 

h 

V 

B 

N^C 

/E 

n 

X 


254  HYDROSTATICS. 

co-ordinates  of  the  point  of  application  C  of  the  resultant  R, 
we  have  (Art.  43) 

R  =  Z.p,  Rx='Zpx,  Ry=2.py. 

If  6  be  the  inclination  of  the  side  of  the  vessel  or  immersed 
surface  to  the  surface  of  the  fluid,  we  have 

h=x  sin.  6. 
Hence  p=gpxk  sin.  6, 

and  R=2. gpxk  sin.  Q=gp  s'm.  O.l.xk, 

Rx=2.gpx"k  sin.  0=gp  sin.  6.2.x*k, 
Ry=2.gpyxk  sin.  d=gp  sin.  d.l.xyk. 
_     2.x"- k 

•••  «S=I3'  <143> 

and  v=Y^k-  (144> 

When  the  upper  boundary  is  below  tne  surface  of  the  fluid 
and  at  a  distance  a  from  it,  then,  since  the  pressures  p  are 
limited  to  the  immersed  surface,  h=(a+x)  sin.  6,  and  we  shah 
have 

_    S(ax+x')k  ,       , 

x=  _\    ,      '  ,  (145) 

_     E(a+x)yk 

y-k+^k-  <I46> 

Further,  if  the  axis  of  x  is  so  taken  that  it  will  bisect  ever} 
horizontal  line  of  the  immersed  surface,  the  pressures  on  oppo 
site  sides  of  this  axis  will  obviously  be  equal,  and  the  center 
of  pressure  will  lie  in  this  axis  or  y=0. 

It  will  be  observed  that  the  numerator  of  (143)  is  the  mo 
ment  of  inertia  of  the  surface,  and  that  the  denominator  .is  the 
statical  moment.     Hence  the  denominator  is  equal  to  the  area 
of  the  surface  multiplied  by  the  depth  of  its  center  of  gravity 
(29).     Hence,  if  A  be  the  area  and  ic,  this  depth, 

_    Z.x*k 
*Tf?  <14?) 

(148) 


»      Ax,  ' 
396.  Prop.  To  find  the  center  of  pressu  re  of  a  rectanguitu 


CENTER     OF     PRESSURE. 


255 


surface  vertically  immersed,  and  having  one  side  in  the  surface 
of  the  fluid. 

Let  YY,  be  the  line  of  intersection 
of  the  immersed  surface  ABCD  with  Y,- 
the  surface  of  the  fluid,  and  conceive 
ABCD  divided  into  n  rectangles  by 
horizontal  lines,  n  being  a  very  large 
number.  Let  AB=&  and  AD=/i, 
and  draw  OX  bisecting  the  rectangle. 
The  center  of  pressure  is  obviously 
in  OX  and  y—Q.     To  find  x,  we  have,  since  the  height  of  each 

h 

small  rectangle  is  -,  for  the  area  of  each, 


,     bh 
n 

The  distances  of  these  rectangles  from  the  surface  are 

h   2/i    3/t    . 

-,  — ,  — ,  &c. ; 
n    n     n 

If    2Vi'    3V*'    B 

— ri  -T7->  &c- 


their  squares, 


n      n 


n 


„     bh3    bh3.2*    M3.32       B 

Hence  2.xk=—r-\ =— H = — K  &c 

n3        n3  n3 


=^(l2+22  +  32  + 


n). 


But  when  n  is  very  large,  the  sum  of  the  mth  powers  of  thft 
natural  numbers  1,2,  3,  &c,  to  n  is 


m+l 


Also, 


.-.  X.x*k=—.--=±bh3. 
n3  3      3 

Axl=bh.\h=\bh\ 

ibh' 


<M* 


-§*> 


or  the  center  of  pressure  is  two  thirds  the  height  of  the  rect- 
angle below  the  surface  of  the  fluid. 

397.  Prop.   To  find  the  center  of  pressure  when  the  immersed 


250 


HYDROSTATICS. 


surface  is  a  triangle,  having  one  side  horizontal  and  the  opposite 
vertex  in  the  surface. 
A       p         b  Let  BDC  be  the  triangle,  h  its  height, 

and  b  its  base,  and  suppose  the  triangle  di- 
vided as  before  into  n  horizontal  divisions. 
The  heights  of  the  successive  divisions 
h 


the  lengths, 
the  depths, 


are  -  ; 
n 

b    2b   3b    c 
-,  — ,  — ,  &c. ; 
71    n     n 

h        _2h       _3h 

n 


n     -      n 

Their  areas,  considering  the  horizontal  sides  of  the  success- 

'•ve  trapezoids  as  differing  insensibly  from  each  other,  are 

_bh       _2bh       _3bh 

1     n*     a      n       3      n 

Multiplying  these  by  the  squares  of  their  depths,  and  taking 

their  sum,  we  have 

„     bh*    23bh*     3sbh*       c  n'bh5 

S.a;-A=— j-+— j — H— 1 — b  &c — — , 

n        n  n  n 

bh5 

=— (l3+23+33+,  &c n3), 

n 

bh'n*       „, 

=—.-=\bh\ 
n    4 

Also,  K=\bh,  and  Sj  =§ A. 

Hence  *=#&=& 

If  now  BE  be  drawn  bisecting  the  base,  the  center  of  press- 
ure will  be  in  BE,  and  at  a  distance  from  the  surface  BO=fBE. 

398.  Prop.  To  find  the  center  of  pressure  when  the  base  of  the 
triangle  lies  in  the  surface. 

In  this  case  the  heights  of  the  successive  divisions  of  the  tri- 
angle ADB  (Fig.,  Art.  397)  are  - ; 

lL    .         .        .     b    ,     2b  .     2b    „ 

the  lengths,    b — ,  b ,  0 ,  &c. : 

&  nun 


CENTER     OF     PRESSURE. 


257 


the  depths,    xl=-,  x2  = — ,  x. 


3h 


n 


n 


n 


,  &c. 


bh    bh          bh    2b  h          bh    3b  h 
the  areas,       k= r,  h,= r>  «3= i~s  &c,> 


n 


n 


which,  multiplied  by  the  squares  of  the  depths,  and  their  sura 
taken,  give 

bh'    bh3    2*bh3    23bh3    3W    3"6/** 

2.z*k=— H i ~ ^ i —+»  &c-> 

n       n        n  n  n  n 

=^(l2+22+32+  . . .  ?i2)-^r(l3+23+33+  ...»•) 

Wn»     M3n4       „,       773 
71s  3       %    4      3  4 


and 


H.x'k- 


±bh.\h=\bh\ 

^blf 


Mi- 


=u. 


B 


Draw  DF  bisecting.  AB,  and  take  FO'=iFD.  0'  is  the  cen- 
ter of  pressure  of  DAB. 

399.  Prop.  To  find  the  center  of  pressure  when  a  rectangle  is 
immersed  vertically,  having  a  side  parallel  to  the  surface  of  the 
fluid  and  at  a  given  distance  below  it. 

Let  a  be  the  distance  of  EF,  the  upper 

side  of  EFCD  from  AB  the  surface  of  the 

fluid,  and  conceive  the  rectangle  divided 

as  before  into  n  divisions  whose  heights 

h        .  bh      TT  .        .       .         . 

are  — ,  and  areas  — .     Using  (145),  and 
n  n 

reckoning  x  from  EF,  the  origin  of  the 

surfaces,  the  depths  are 

h         2h         3h   „ 
a-\ — ,  a-\ — ,  a-\ — ,  &c. 
n  n  n 

v    ,     abh"    bh3    2abh*    2*bh*    3abh*    3*bh3 
Hence  2.{a+x)xk=-^~\  -7^-+— 3 —    ~Zi —   ~ir~H — n — h&c- 


E 


D 


M 


n 
abh 


bh3 


— r(l+2+3+....n)+— (l2+22+32+...ns) 
ablf  ?i2    bh  n* 

R 


258 


HYDROSTATICS. 


A.nd         Z(a+x)k=Kxl  =  bh(a-\-\h)  =  abh-\-\bh'i 


__\abh*+\bh'i     ]  2h+3a_ 
h  h+2a  = 
d3-a3 


abh+±bh* 

,  2A+3a 
j^ft.  ,  .  _ — l-a=f. 


MO. 


/j+2a 


3'cf-a'J' 


Hence  LO=a+x: 

in  which  d=AD=a+A. 

400.  Prop.    To   determine  the  conditions  of  equilibrium  oj 
fluids  of  different  density. 

When  fluids  of  different  densities  which 
do  not  mix  or  unite  chemically  are  contain- 
ed in  the  same  vessel  they  will  arrange  them- 
selves in  the  order  of  their  densities,  the  most 
dense  taking  the  lowest  position.  This  is  ob- 
vious from  the  ready  displacement  of  their 
particles  and  Art.  390,  Cor.  1. 

The  limiting  surfaces  of  each  will  likewise  be  horizontal ; 
for,  in  order  to  equilibrium,  any  horizontal  stratum  must  sus- 
tain in  every  part  of  it  the  same  pressure.  But  if  the  common 
surface,  as  EF,  of  two  of  them  be  inclined,  then  a  horizontal 
stratum,  as  HR,  will  sustain  at  different  points,  as  O  and  O', 
superincumbent  columns  of  fluid  of  different  weights,  and  the 
equilibrium  will  not  subsist. 

401.  Prop.  In  communicating  tubes,  the 
heights  at  which  fluids  of  different  densities 
will  stand  above  their  common  base  when  in 
equilibrium,  are  inversely  as  their  densities. 
In  the  bent  tube  AED,  let  one  fluid  oc- 
cupy the  portion  AR,  and  the  other  the  por- 
tion RED,  the  fluids  having  the  common 
base  HR.  Let  h  be  the  height  of  the  fluid 
in  AR  above  HR,  and  p  its  density,  h'  the 
height  of  the  surface  CD  above  HR,  p'  its 
density,  and  let  the  common  base  HR=A. 
The  pressure  of  the  fluid  in  AR  on  A  is 
gphA,  and  that  of  the  fluid  in  RED  on  the 
same  base  gp'h'A.  In  equilibrium  these 
pressures  will  be  equal. 


H 


B 


D 


EXAMPLES.  259 

Therefore  gphA=gp'h'A, 

*     P' 
and  77=-.  (149) 

«      P 

402      EXAMPLES. 

jG'.r.  1.  A  cubical  vessel,  each  side  of  which  is  ten  feet  square. 
is  filled  with  water,  and  a  tube  thirty-two  feet  long  is  fitted  to 
an  aperture  in  it,  whose  area  is  one  square  inch.  If  the  tube 
be  vertical,  of  the  same  diameter  as  the  aperture,  and  filled 
with  water,  what  is  the  pressure  on  the  interior  surface  of  the 
vessel,  neglecting  the  weight  of  the  water  it  contains  ? 

Since  the  weight  of  a  cubic  foot  of  water  is  1000  oz.,  the 

•     i  r  ,  .       .       ,      .       1000  ,      .  .     . 

weight  ot  one  cubic  inch  is  ,  „  ,=.5787  oz.,  and  the  weight 

1728 

of  32X12=384  cubic   inches   is  384X. 5787=222.2208  oz.  = 

13.8888  lbs.     This  is  the  pressure  on  the  aperture,  or  one 

square  inch  of  the  surface  of  the  water  in  the  vessel* 

The  number  of  square  inches  in  the  interior  surface  of  the 
vessel  is  6X  102X  144=86,400. 

Then,  Art.  381,  the  pressures  being  in  the  ratio  of  the  areas 
pressed,  the  whole  pressure  on  the  interior  surface  will  be 
P=13.888SX86399=1,200,000  lbs.  nearly. 

Ex.  2  What  is  the  pressure  on  the  bottom  of  the  vessel  in 
Ex.  1  when  the  weight  of  the  water  in  the  vessel  is  taken  into 
account ;   1°,  without  the  vertical  tube,  and  2°,  with  it. 

1°.  The  base  of  the  vessel  is  100  square  feet,  and  its  depth 
10  feet.     Hence,  Art.  385, 

P,  =  10X  100X62.5=62,500  lbs.  =  the  weight  of  the  water. 

2°.  The  pressure  transmitted  to  each  square  inch  of  the  base 
by  the  water  in  the  tube  is  13.8888  lbs.  Hence  the  whole 
transmitted  pressure  P2  =  13.8888  X  102X  144  =  200,000  lbs. 
nearly,  and  the  entire  pressure  on  the  base  is  P=262,500  lbs. 

Ex.  3.  What  is  the  pressure  on  each  vertical  side  of  the  ves- 
sel without  the  tube  ? 

The  depth  of  the  center  of  gravity  of  each  side  of  the  vessel 
is  5  feet.     Hence,  Art.  386, 

P3  =  5X102X62.5  =  31,250  lbs., 
equal  one  half  the  pressure  on  the  base. 


260  HYDROSTATICS. 

Thus  we  have  the  whole  pressure  on  the  interior  surface  of 
the  vessel,  including  that  produced  by  the  water  in  the  tube, 

P+P,+4P3  =  1,200,000 
+      62,500 
+    125,000 
=  1,387,500  lbs. 
Ex.  4.  Required  the  pressure  on  the  base  of  a  conical  ves- 
sel  filled  with  water,  the  radius  of  the  base  being  r=5  feet,  and 
the  altitude  a=10  feet. 
By  (137),  P=62.5AA  lbs.,  in  which  h=a,  and  A=nr\ 

.-.  P=62.5:rr«=62.5X3.1416X25X  10=49,087.5  lbs. 
Ex.  5.  Required  the  normal  pressure  on  the  concave  surface 
of  the  cone  of  Ex.  4. 


The  slant  height  of  the  cone  is  Vcf+r*,  and  the  concave 
surface  is  2m .     The  vertical  depth  ot  the  center  of 

gravity  of  the  concave  surface  is  — . 

Hence (137),  Px  =  62.5X^*  vV+7 

o 

=62.5XfXl0X3.1416X5Vl25=73,175.11bs. 

Ex.  6.  Required  the  vertical  pressure  on  the  concave  sur- 
face of  the  same  cone. 

Since  the  normal  to  the  side  of  the  cone  makes  the  same  an- 
gle with  the  axis  of  the  cone  that  the  side  of  the  cone  makes 
with  the  base,  the  cosine  of  the  inclination  of  the  normal  to  the 


axis  is 


vV+ra 
Hence  the  vertical  pressure  {Art.  387) 

Pa=P.  -=^==62.5Xfa:rrvV+r.-=^==62.5Xf7rra, 
vV+ra  vV+r3 

=|P  {Ex.  4)  =  32,725  lbs. 

Ex.  7.  Required  the  resultant  of  all  the  pressures  on  the  in- 
terior surface  of  the  same  cone. 

By  Art.  388,  the  resultant  of  the  horizontal  pressures  is  zero 


EXAMPLES.  261 

and  the  resultant  of  the  vertical  pressures  is  equal  to  the  ex- 
cess of  the  downward  pressure  over  the  upward  pressure. 

.-.  R  =  P  (Ex.  4)-Pa  (Ex.  6)  =  16,362.5  lbs., 
and  this  is  the  weight  of  the  water  in  the  cone. 

Ex.  8.  A  rectangular  parallelogram,  whose  sides  a  and  b  are 
26  feet  and  14  feet  respectively,  is  immersed  in  water  with 
the  side  b  in  the  surface,  and  is  inclined  to  the  surface  at  an 
angle  </>=56°.35'.  Required  the  pressures  Pj  and  P2  on  the 
parts  into  which  the  parallelogram  is  divided  by  its  diagonal. 

Ans.  P,=   82286.5  lbs. 
P2  =  164573.0  lbs. 

Ex.  9.  When  a=30  feet,  6  =  20  feet,  and  0=59°.38',  re- 
quired the  pressures  on  the  equal  parts  into  which  the  paral- 
lelogram is  divided  by  a  line  parallel  to  the  horizon. 

Ans.  P,  =  121332|  lbs. 
P2  =363998    lbs. 

Ex.  10.  When  the  parallelogram  of  Ex.  9  is  vertical,  how 
far  from  the  surface  must  the  dividing  line  be  drawn  that  the 
pressures  on  the  two  parts  may  be  equal  ? 

Ans.  z=21.213  feel. 

Ex.  \\.  A  sphere  10  feet  in  diameter  is  filled  with  water. 
Required  the  entire  pressure  on  the  interior  surface,  and  the 
weight  of  the  water. 

Ex.  12.  A  vessel  in  the  form  of  a  paraboloid,  with  vertex 
downward,  is  filled  with  water,  and  revolves  uniformly  on  its 
axis.  Required  the  time  of  one  revolution  when  the  angular 
velocity  is  just  sufficient  to  empty  it. 

Ex.  13.  The  concave  surface  of  a  cylinder  filled  with  fluid 
is  divided  by  horizontal  sections  into  n  annuli  in  such  a  man- 
ner that  the  pressure  on  each  annulus  is  equal  to  the  pressure 
on  the  base.  Given  the  radius  of  the  cylinder.  Required  its 
height,  and  the  breadth  of  the  mth  annulus. 

Ex.  14.  A  regular  tetrahedron,  with  one  face  horizontal,  is 
filled  with  water.  Compare  the  pressures  on  the  base,  and  on 
the  other  sides  with  the  weight  of  the  water. 


262 


H  V  I)  R  n  s  T  A  T  I  C  S. 


¥r? 


Ex.  15.  An  irjn  vessel  40  feet  long,  every  transverse  sec- 
tion of  which  is  an  isosceles  triangle  whose  base  is  16  feet  and 
altitude  20  feet,  floats  with  its  vertex  downward.  If  a  cubic 
foot  of  iron  weighs  487.5  lbs.,  required  the  depth  to  which  the 
vessel  will  sink  when  the  sides  and  ends  are  one  quarter  of  an 
inch  thick  and  there  is  no  deck. 

Ex.  16.  An  embankment  of  brick- work, 
of  which  ABC  is  a  section,  weighs  120 
lbs.  to  the  cubic  foot.  Its  height  AB  is 
14  feet,  and  its  base  BC  is  6  feet.  Find 
whether  or  not  the  embankment  will  be 
overthrown  by  the  pressure  of  water  on 
the  surface  AB. 
ND   B  Draw    AN    bisecting    CB,    and    take 

NG=|NA.  G  will  be  the  center  of  gravity  of  the  section 
and  the  weight  W  will  act  at  G.  Take  AM=|AB,  and  M 
will  be  the  center  of  pressure  (Art.  396).  The  resultant  P  of 
the  pressure  of  the  water  will  act  at  M  and  will  pass  through 
G.  The  moment  of  P  to  turn  the  embankment  over  C  will  be 
P.CO,  while  the  moment  of  W  to  resist  the  overturn  will  be 
W.CD.  If  P.CO> W.CD,  the  embankment  will  be  overturned. 
Since  the  embankment  is  uniform  throughout  its  length,  as 
also  the  pressure  on  it,  we  may  determine  the  stability  by 
taking  one  foot  in  length. 

Now  W=1X14X6X|X120=5040  lbs., 

and  P=lXl4X7X62.5=6125lbs., 

OC  =  BM  =  ixl4  and  CD  =  f  X6=4. 
.•-.  W.CD=20160, 
and  P.CO=28583i. 

The  latter  being  greater  than  the  former,  the  embankment 
will  be  overturned. 

Ex.  17.  A  wall  of  masonry,  a  section  of  which  is  a  rectan- 
gle, is  10  feet  high,  3  feet  thick,  and  each  cubic  foot  weighs 
100  lbs.  Find  the  greatest  height  of  water  it  will  sustain 
without  being  overturned. 

Ex.  18.  If  the  height  of  the  wall  be  8  feet,  its  thickness  6 
feet,  and  each  cubic  foot  weigh  180  lbs.,  find  whether  it  will 
stand  or  fall  when  the  water  is  on  a  level  with  the  top. 


CHAPTER  II. 


SPECIFIC     GRAVITY. 


403.  Def.  The  specific  gravity  of  a  body  is  the  ratio  of  the 
weight  of  the  body  to  the  weight  of  an  equal  volume  of  some 
other  body  taken  as  the  standard  of  comparison,  and  whose 
specific  gravity,  therefore,  is  taken  as  the  unit. 

Water  is  generally  employed  as  the  standard  of  comparison 
for  solids  and  liquids,  and  atmospheric  air  for  aeriform  fluids. 

Cor.  If  u,  p,  o,  and  w  be  the  volume,  density,  specific  gravity, 
and  weight  respectively  of  one  body,  and  vp  p,,  o1,  w^  the 
same  of  another  body;  then,  since  v=i\, 
o       w       g-p-v       p 

—=—=-*-*— =r~,  (150) 

°,      w,      g.pxV,      p, 
or  the  ratio  of  the  specific  gravities  of  two  bodies  is  equal  to 
that  of  their  densities. 

404.  Prop.  To  find  the  specific  gravity  of  a  body  more  dense 
than  water. 

If  the  body  be  immersed  in  water  it  will  descend  (Cor.,  Art. 
390).  Let  w  be  the  absolute  weight  of  the  body,  and  w1  its 
weight  or  the  force  with  which  it  will  descend  when  immersed 
in  water.  Then  the  loss  of  weight  in  water  is  w  —  wJf  and  this 
is  equal  to  the  upward  pressure  of  the  water,  or  to  the  weight 
of  a  volume  of  water  equal  to  that  of  the  solid.     Hence 

w         absolute  weight 

o= =- — ^— .  (151) 

w  —  wx       loss  ot  weight  v 

The  absolute  weight  as  well  as  the  loss  of  weight  is  ascer- 
tained by  the  hydrostatic  balance,  which  differs  from  the  or- 
dinary balance  only  in  having  a  hook  appended  beneath  one 
of  the  scale  pans,  to  which  the  body  may  be  suspended  by  a 
fine  thread  and  allowed  to  sink  in  a  vessel  of  water  beneath  it. 
The  body  is  first  placed  in  the  scale  pan  and  counterpoised  by 


Z(f4  HYDRO&TATICS 

a  weight  w,  and  then  suspended  to  the  hook,  and  when  irn 
mersed  in  the  water  its  counterpoise  w1  again  determined. 

405.  Prop.  To  find  the  specific  gravity  of  a  body  less  dense 
than  water. 

Since  the  body  A  is  less  dense  than  water,  it  will  not  de- 
scend in  the  water  by  its  own  gravity.  Let  a  more  dense 
body  B  be  attached  to  it,  and  call  the  compound  body  C. 

Let  iv  =  absolute  weight  of  A, 

w'  =       "  "        "  B,  and  w\=  its  weight  in  water 

w"=       "  "        "  C,and  w'[=  "         "      "       " 

Then  w'  —  w\  =loss  of  weight  of  B, 

w"—w"l=  "     "        "       "  C, 

and  (w"  —  w'()  —  (w'~ w\)  —  \oss  of  weight  of  A,  the  upward 
pressure  of  the  water  on  A,  and  therefore  equal  to  the  weight 
of  a  volume  of  water  the  same  as  that  of  A. 

But  w"—w'+w. 

and,  by  substitution, 

(w"— w")  —  {w'— w'1)=w'+w— w"— w'+w\=w+w\  —to','. 

.-.  «t=-— jj.  (152) 

w-\-wl—w{  v 

Hence,  add  to  the  absolute  weight  of  the  body  the  difference 
of  the  weights  of  the  more  dense  and  compound  bodies  in 
water,  and  divide  the  absolute  weight  of  the  body  by  the  sum. 

406.  Prop.   To  find  the  specific  gravity  of  a  liquid. 

Let  a  body  whose  weight  is  iv  be  weighed  both  in  water 
and  in  the  liquid,  the  weight  in  the  former  being  w , ,  and  in  the 
latter  w2.  Then  w— to,  and  io—w2  are  the  weights  of  equal 
volumes  of  the  two  liquids.     Hence 

w—w„ 
o= 2-\  (153) 

w  —  ivl 

or,  if  an  empty  bottle  whose  weight  is  w,  weighs  when  filled 
with  water  wl,  and  when  filled  with  the  liquid  w2,  then 

iv  n— 10 
c=-^ .  (154) 


SPECIFIC     GRAVITY.  265 

407.  Prop.  To  find  the  weights  of  the  constituents  in  a  me- 
chanical composition  when  the  specific  gravities  of  the  compound., 
and  constituents  are  known. 

Let  w,  to,,  w2  be  the  weights  of  the  compound  and  constitu- 
ents respectively. 

a,  a,,  a2  their  respective  specific  gravities, 
v,  u,,  v2  their  volumes. 
In  all  merely  mechanical  combinations, 

v  =  vt  +v2  (a),  and  w=wl  +w2  (b). 

t>  w  w.  .  w„ 

but  v= — ,  v= ,  and  u  = . 

gP  gPi  gP2 

,   .     W      10,       w0 

•'.  (a)  -=—+^-; 
P      Pi      P2 

or,  since  their  densities  are  as  their  specific  gravities  (150), 

w    vk     wn 
-=  —  +—. 

°         Gl  °2 

Substituting  in  succession  the  values  of  wl  and  w2,  obtained 
from  (b).  we  have 


10, 


(\      1\      /l        n      (on-a)a. 


IV, 


.=W(I_IW±_1)=£^%„.    (I56) 

\o     o2J       \a2     olJ      (cr,—  o2)a  x       ' 

408.  Hydrometer.  Instruments  for  determining  the  specific 
gravity  of  fluids  are  called  Hydrometers  or  Areometers.  They 
are  made  of  glass,  brass,  &c,  and  are  of  two  kinds :  one  in 
which  the  weight  is  constant,  the  other  in  which  the  volume  is 
constant.  In  its  simplest  form  the  areometer  consists  of  a  hol- 
low globe,  one  of  whose  diameters  is  prolonged  in  a  flat  oi 
cylindrical  stem  of  uniform  size,  and  to  the  other  extremity  is 
attached  a  smaller  globe  loaded  with  mercury  or  shot,  that  it 
may  float  in  a  vertical  position. 

1°.  Areometer  of  a  constant  weight. 

The  annexed  figure  may  represent  this  form  of  the  instru 
ment.  To  graduate  the  stein,  suppose  it  to  sink  in  distilled 
water,  at  a  given  temperature,  to  a  point  s,  and  let  the  distance 
sr  be  divided  into  any  number  of  equal  parts,  and  continued 


200 


HYDROSTATICS. 


upward  from  s.  When  immersed  in  another  fluid 
suppose  it  to  sink  to  t,  distant  from  s,  x  divisions.  Let 
V  be  the  volume  of  the  portions  immersed  in  water, 
v  the  volume  included  between  any  two  divisions  of 
the  stem,  and  o  the  specific  gravity  of  the  second  fluid. 
Then  p  and  px  being  the  densities  of  this  fluid  and 
water,  the  weight  of  the  water  displaced  by  the  in- 
strument will  be  gP-tV,  the  weight  of  the  other  fluid 
displaced  will  be  gp(V—vx),  and,  since  each  is  equal 
to  the  weight  of  the  instrument, 

gP1V=gp(V-vx). 
V 

o         V  v 

(a) 


a=- 


V-vx     V 


If  c  be  previously  known,  and  x  be  observed,  we  can  de- 
,'mine  the  value  of 

V 


— =x.- . 

v         \—a 


V 

Substituting  in  (a)  this  value  of—,  and  putting  for  x,  1,2,  3 

&c,  the  corresponding  values  of  a  for  each  division  of  the  scale 
will  be  known.     These  may  be  marked  on  the  instrument,  or 
the  divisions  may  be  numbered,  and  a  table  of  their  values 
formed  to  accompany  the  instrument. 
Cor.  From  (a)  we  have 

v,    r 

x-^-(l 

Giving  to  a  a  small  increment  do,  the  corresponding  increment 
of  a;  is 

v  o 
that  is,  for  any  small  increment  of  the  specific  gravity  of  the 
fluid  the  corresponding  change  in  the  depth  of  immersion  of 

V  1 

the  instrument  varies  as  — .— ,  which  may  be  considered  a  meas- 


ure  of  the  susceptibility  of  the  instrument. 


STECIFIC     G  E  A  V  1  T  Y. 


26" 


2J.  Areometer  of  a  constant  volume. 

The  principal  obstacle  to  the  use  of  the  areometer  of  a  con- 
stant weight  is  the  inconvenience  and  difficulty  of  calculating 
and  marking  against  the  different  divisions  of  the  stem  of  each 
instrument  a  different  scale  of  specific  gravity,  and  construct- 
ing the  stem  of  that  perfectly  uniform  thickness  which  is  nec- 
essary to  the  accuracy  of  the  observations. 

To  obviate  these  difficulties,  Fahrenheit  conceived  the  idea 
of  sinking  the  instrument  always  to  the  same  depth  by  means 
of  weights  to  be  placed  in  a  cup  at  the  end  of  the  stem. 

Let  W  be  the  weight  of  the  instrument,  wY  and  w  the  weights 
respectively  necessary  to  sink  it  to  the  same  point 
a  of  the  stem  in  water,  and  in  the  fluid  whose  ^ 
specific   gravity   a   is  required,  V  the   constant 
volume  of  the  portion  immersed,  p1  and  p  the 
densities  of  the  writer  and  the  fluid.     Then 
gpY=Yi+w  and  gp1Y=W+w1. 
_  p  _  W+w 
~~  p~  W+ttJ,' 
Cor.  Differentiating,  we  get 

dw={W +w  1)do=gp  t  Ydo ; 
that  is,  for  any  given  small  variation  do  in  the 
specific  gravity  of  the  fluid,  the  variation  of  w  is  as  V,  or  the 
susceptibility  of  the  instrument  is  as  the  volume  of  the  portion 
immersed. 

409.  Nicholson's  Hydrometer.     This  is  a  modification  of  the 

preceding,  to  adapt  it  to  the  determination        

of  the  specific  gravity  of  solids  as  well  as  t  yy 
liquids.  For  this  purpose,  a  metallic  basket 
is  attached  to  the  lower  extremity,  in  which 
the  body  may  be  placed,  and  its  weight  in 
water  ascertained.  The  basket  admits  of 
reversal,  so  that  the  body  may  be  retained 
under  water  when  specifically  lighter. 

Let  w  be  the  weight  in  C  necessary  to 
sink  the  instrument  to/.  Replacing  w  by 
the  body,  let  w,  be  the  weight  which  must 


26*3  HYDROSTATICS. 

be  added  to  sink  the  instrument  to  the  same  depth/.  Removing 
the  body  to  the  basket  beneath,  let  w2  be  the  weight  in  C  req- 
uisite to  sink  the  instrument  a  third  time  tof. 

Calling  W  the  weight  of  the  instrument,  Wx  that  of  the 
water  displaced  by  it  when  immersed  to  f,  V  the  volume  of 
the  body,  p  its  density,  that  of  water  being  p,. 

Then  W+w=Wl,  (a) 

gpV+W+w^W,,  (b) 

and  gpV+W+w2=Wi+gpiy.  (c) 

Subtracting  (a)  from  (b),  and  (b)  from  (c), 
gpV=w— w,, 

gply=w2-w1. 

p       w—w. 


a=- 


p,     w2—w1 

410.    EXAMPLES. 

Ex.  1.  A  piece  of  wood  weighs  12  lbs.,  and  when  annexed 
to  22  lbs.  of  lead,  and  immersed  in  water,  the  whole  weighs 
8  lbs.  The  specific  gravity  of  lead  being  11,  required  that  of 
the  wood. 

The  specific  gravity  of  lead  being  11,  if  v  be  its  volume, 
llu=22  or  u=2.  But  this  is  the  volume  of  water  displaced 
by  the  lead,  and  the  specific  gravity  of  water  being  1,  its 
weight  will  be  2.  Therefore  the  loss  of  weight  in  water  is 
2  lbs.,  and  the  actual  weight  in  water  20  lbs. 

Hence,  Art.  405,  w=12,  m/,'=8,  and  w\=20. 
_  w         _         12  1 

°~ w+w\  —w"x  ~  12+20— 8~2' 

Ex.  2.  A  diamond  ring  weighs  69i  grains,  and  when  weigh- 
ed in  water  G4|  grains.  The  specific  gravity  of  gold  being  161, 
and  that  of  diamond  3^,  what  is  the  weight  of  the  diamond  ? 

Let  v,  v'  be  the  volumes  of  the  gold  and  diamond  respective- 
ly, and  /I  the  weight  of  a  unit  of  volume  of  water.  Then  u/l  is 
the  weight  of  a  volume  of  water  equal  in  bulk  to  the  gold,  and 
333tvt  the  weight  of  the  gold.  In  the  same  manner,  |u'A  is  the 
weight  of  the  diamond.     Hence 


SPECIFIC     GRAVITY.  269 

69±=yi>A.+|ij'A,  (a) 

and  64i=:yj?;A+^,A-(y+y')A.  (&) 

Subtracting  (b)  from  (a),  we  have 
5=Ay+Au'. 
.-.  from  (a),  69i=y(5-A»')+|A«/f 

139=165-26Au', 
Au'=l,  fAy'=3i, 
or  the  weight  of  the  diamond  is  three  and  a  half  grains. 

Ex.  3.  A  body  A  weighs  10  grains  in  water,  and  a  body  B 
weighs  14  grains  in  air,  and  A  and  B  connected  together  weigh 
7  grains  in  water.  The  specific  gravity  of  air  being  .0013, 
required  the  specific  gravity  of  B,  and  the  number  of  grains 
of  water  equal  to  it  in  bulk. 

Let  A',  A"  be  the  number  of  grains  in  the  volumes  of  water 
equal  to  the  volumes  of  A  and  B  respectively,  and  a',  a"  their 
specific  gravities.     Then,  by  the  conditions  of  the  question, 

(<j'-l)A'=10,  (a) 

(o"-.0013)A"=14,  (b) 

(a'-l)A'  +  (a"-l)A"=7.  (c) 

From  (a),  (b),  and  (c),  we  have 

,„     14(o"- 1) 

10 -] -=1 

a"-.0013       ' 

14(1-ct")  =  3(<t"-.0013). 

.-.  <t"=.8237. 
Hence,  also,  from  (b), 

A"=0-^oT3=^i=17-OS3srains- 

Ex.  4.  When  73  parts  by  weight  of  sulphuric  acid,  the  spe 
cific  gravity  of  which  is  1.8485,  are  mixed  with  27  parts  of 
water,  the  resulting  dilute  acid  has  a  specific  gravity  equal  to 
1.6321.     Required  the  amount  of  condensation  which  takes 
place  by  the  mixture. 

Let  A'  be  the  number  of  parts  by  weight  in  a  quantity  of 
water  equal  in  volume  to  that  of  the  sulphuric  acid,  and  A"  in 
a  quantity  of  water  equal  in  volume  to  that  of  the  mixture. 

Then  1.8485A'=73, 

or  A' =  39.49 15, 


270  HYDROSTATICS. 

and  1.632U"=73+27=100 

or  A"=61.2707. 

Now  the  condensation  of  the  mixture  will  be  expressed  by  the 
ratio  of  the  diminution  of  the  volume  of  water  and  acid,  when 
mixed,  to  their  united  volume  before  mixture. 

A' +27  -X" 


Hence  condensation: 


A' +27 
5.2208 


:=0.0785. 


66.4915 
Ex.  5.  Two  fluids,  the  volumes  of  which  are  v  and  »',  and 

specific  gravities  o  and  a1,  on  being  mixed,  contract  -th  part  of 

the  sum  of  their  volumes  by  mutual  penetration.     Required 
the  specific  gravity  of  the  mixture. 

Let  a"  be  the  specific  gravity  of  the  mixture ;  then  the  sum 
of  their  weight  before  and  after  mixture  will  be  equal. 

Hence  (vo+v'o')X=(l-  \v+v')o"l, 

.  n      va+v'o' 

which  gives  a"— -. — - — — . 

&  71—1      v+v' 

Ex.  6.  A  body  whose  weight  in  a  vacuum  is  73.29  grains 
loses  24.43  grains  by  immersion  in  water.  Required  its  spe- 
cific gravity.  Ans.  <7=3. 

Ex.  7.  Required  the  specific  gravity  of  a  body  which  weighs 
65  grains  in  a  vacuum,  and  44  grains  in  water. 

Ans.  <r=3.0952. 

Ex.  8.  An  areometer  sinks  to  a  certain  depth  in  a  fluid  whose 
specific  gravity  is  .8,  and  when  loaded  with  60  grains  it  sinks 
to  the  same  depth  in  water.  What  is  the  weight  of  the  in- 
strument? Ans.  w=240  grains. 

Ex.  9.  A  compound  of  gold  and  silver  weighing  i^=  10  lbs. 
has  a  specific  gravity  o=14,  that  of  gold  being  o'=l9.3,  and 
that  of  silver  being  o"=10.5.  Required  the  weight  w'  and  w' 
of  the  gold  and  silver  in  the  compounc 

Ans.  w'= 5.483, 
w"=4.517 


CHAPTER   III. 


COMPRESSIBLE    OR    AERIFORM    FLUIDS. 

411.  Prop.  To  find  the  tension  of  the  atmosphere  or  cthet 
compressible  fluid. 

Let  a  glass  tube  AB,  open  at  one  end  and  A 

closed  at  the  other,  be  filled  with  mercury. 
Retain  the  mercury  in  the  tube  by  the  press- 
ure of  the  finger,  let  it  be  inverted  and  the 
open  end  immersed  beneath  the  surface  of  the 
mercury  in  the  vessel  CDFE.  It  will  now  be 
found  that  a  column  of  mercury,  as  BL,  will 
occupy  a  portion  of  the  tube,  while  the  re- 
maining portion  AL  will  be  void.  This  col- 
umn of  mercury  is  sustained  in  the  tube  by  the 
pressure  of  the  atmosphere  on  the  surface  of 
the  mercury  in  the  cistern  CF,  and  as  there  is 
no  pressure  on  the  mercury  at  L,  the  equilib- 
rium is  due  to  the  equality  of  pressures  of  the 
atmosphere  and  mercury  on  the  common  base 
B  (Art.  401).  If  we  now  suppose  the  cistern 
covered,  so  as  to  separate  the  air  without  from 
that  within  the  cistern,  the  pressure  of  the  ex- 
ternal air  can  not  be  communicated  to  that 
within,  and  the  mercurial  column  must  be  sustained  by  the  ex- 
pansive force  or  tension  of  the  inclosed  air,  and  be  a  measure 
of  it. 

If  the  tension  of  any  other  elastic  fluid  inclosed  in  a  vessel 
be  required,  let  the  tube  I  from  the  cistern  CEFD  be  fitted  to 
an  aperture  in  the  vessel,  and  a  communication  be  thus  estab- 
lished between  the  inclosed  fluid  and  the  mercury  in  the  cis- 
tern. The  mercury  in  the  tube  will  then  rise  or  fall  till  an 
equilibrium  takes  place  between  the  expansive  force  of  the  fluid 


272  HYDROSTATICS. 

and  the  weight  of  the  mercurial  column.  The  height  at  whicn 
tne  mercury  stands  in  the  tube  above  the  surface  of  mercury 
in  the  cistern  is  ascertained  by  a  graduated  scale  attached  to 
the  tube. 

412.  Schol.  The  mean  pressure  of  the  atmosphere  at  or 
near  the  level  of  the  sea  is  generally  employed  in  mechanics 
as  the  unit  of  pressure,  and  other  expansive  forces  are  com- 
pared with  this  and  expressed  in  atmospheres.  It  has  been 
ascertained  by  the  barometer  that  the  mean  pressure  of  the 
atmosphere,  at  a  temperature  of  50°,  is  equivalent  to  a  column 
of  mercury  30  inches  in  height ;  or,  the  specific  gravity  of  mer- 
cury being  13.598,  to  a  column  of  water  13.598X2.5  feet=34 
feet ;  or,  the  specific  gravity  of  air,  at  a  temperature  of  50°, 

being  — — ,  to  a  column  of  air  of  uniform  density,  27,540  feet 

=  5.2  miles  in  height. 

The  tension  is  also  measured  by  the  pressure  of  the  atmos- 
phere on  a  unit  of  surface.     Now  30  cubic  inches  of  mercury 

is  equal  to  13.598X30  cubic  inches  of  water  =407.94 Xr-r; 
1  1728 

cubic  feet  of  water= 0.23607 X 1000  oz.=  14.75  lbs.  on  the 

square  inch. 

The  instrument  above  described  involves  the  essential  parts 

of  a  barometer,  a  more  particular  description  of  which  belongs 

to  Physics.     Generally,  instruments  employed  to  determine  the 

tension  of  elastic  fluids  are  called  manometers. 

413.  Prop.  To  show  that  the  tension  of  an  aeriform  fluid  is 
inversely  as  its  volume. 

Let  AB  be  a  vertical  tube  which  communicates  with  the 
cylindrical  vessel  DEOjH,,  closed  at  the  top,  and  mercury  be 
carefully  introduced  into  the  tube,  so  as  to  fill  the  horizontal 
part  HjOjBC,  and  leave  the  air  in  DOa  of  the  same  density 
as  the  exterior  air.  The  inclosed  air  will  then  have  a  tension 
of  one  atmosphere.  Take  DH  =iDH,,  DH3  =  iDHj,  &c, 
and  draw  the  lines  H202R2,  H303R3,  &c.  Now  if  mercury 
be  poured  into  the  tube  AB  until  it  stands  at  H202  in  the 
cylinder,  it  will  be  found  to  stand  in  the  tube  at  the  height 


COMPRESSIBLE  OR  AERIFORM  FLUIDS. 


273 


L2R2  =  30  inches  above  H202;  or,  when  the  air 
in  the  cylinder  is  reduced  to  half  its  volume,  :ts 
tension  or  expansive  force  is  two  atmospheres. 
When  the  mercury  rises  in  the  cylinder  to  the 
height  H3O3,  it  will  be  found  to  stand  at  the 
height  L3R3  =  60  inches  above  H303;  or,  when 
the  air  is  reduced  to  one  third  of  its  volume,  it 
has  a  tension  of  three  atmospheres,  or,  general- 
ly, the  tension  is  inversely  as  the  volume.  The 
truth  of  the  law  has  been  tested  experimentally 
as  far  as  26  atmospheres,  and  for  all  fractions 
of  an  atmosphere. 

If  V  and  V  be  the  volumes  of  a  given  mass  h* 
of  air,  p  and  p'  the  corresponding  tensions  or  Hz 
pressures  on  a  unit  of  surface,  then 


D 

m 


E 


^-,=^r,orpV=p'Y'. 


V 


V 


(157) 


U 


L2 


R4 


Ri 


B 


This  law  is  called  the  law  of  Mariotte,  from  the  discoverer 

414.  Cor.  Let  p  and  p'  be  the  densities  of  any  mass  of  an 
corresponding  to  the  volumes  V  and  V.  Then,  since  the 
density  is  inversely  as  the  volume, 

p     V     p 

p7=V=^  <158> 

or  the  elastic  force  is  directly  as  the  density. 

415.  PRor.  To  estimate  the  effect  of  heat  on  the  volume  and 
tension  of  atmospheric  air. 

When  air  is  inclosed  in  a  vessel  and  heat  is  applied,  its  elastic 
force  is  increased,  as  may  be  shown  by  the  method  indicated 
in  ArL  411,  and  by  the  same  method  the  increase  in  the  ten- 
sion for  a  given  increase  of  temperature  may  be  ascertained. 
Experiment  indicates  that  the  tension  of  a  given  volume  of 
dry  air  increases,  by  being  heated  from  the  freezing  to  the 
boiling  point,  0.367  of  its  original  value,  and  therefore,  if  the 
tension  remains  the  same,  the  volume  will  increase  36.7  pei 
cent.  Let  v  be  the  increment  of  volume,  the  tension  remain- 
ing the  same,  for  one  degree  of  Fahrenheit's  thermometer ;  then 

S 


274  HYDROSTATICS. 

v=0.002039,  and  for  an  increase  of  *  degrees  of  temperaturev 
the  increase  of  volume  will  be  r*=0.002039*. 

If  now  V0  be  the  original  volume  of  a  given  mass  of  air,  and 
it  be  heated  *,  degrees,  the  tension  remaining  the  same,  the 
new  volume  V,  will  be 

V,=(1+,*,)V0; 
•md  when  heated  *„  degrees  the  corresponding  volume  will  be 

Ya=(l+vt2)Y0. 

V,      1+v*,      1+0.002309*, 


"  Y2~l+vt2      1 +0.002309*/ 

But  the  densities  p,  and  p2  are  inversely  as  the  volumes. 

P,     V2     l+v*2 


Pa     V,     l+vt,' 

If,  also,  a  change  take  place  in  the  tensions  at  the  same  time, 
letp0be  the  tension  at  32°,/?,  the  tension  at  32°+*,,andp2  that 
at  32°+*2.     Then,  since  the  tension  is  inversely  as  the  volume, 

V,  =  (1+,*,)^V0  and  V2  =  (1+.*2)^V0. 
"Y2     (l+vt2)Pl' 

and         pT~(i+^-?:~(T+^)v       (160) 

in  which  &,  and  b2  represent  the  measures  of  the  tensions/?, 
and  p2,  or  the  corresponding  heights  of  the  barometer. 

416.  Prop.  To  find  the  density  of  the  air  at  different  tempera- 
tures and  under  different  pressures. 

By  accurate  experiments  the  weight  of  a  cubic  foot  of  air  at 
a  temperature  of  32°,  when  the  barometer  stands  at  30  inches, 
is  found  to  be  p=0.081 12  lbs.  avoirdupois.  Hence,  for  the  tem- 
perature 32°+*°. 

p  0.08112       „  .,„,. 

^  =T+^Tl  +0.002039*  IbS'  (161) 

If,  also,  the  barometer,  instead  of  &=30  inches,  should  stand 
at  some  other  height,  as  bl  inches,  the  density  will  be  express- 
ed by 

p     &,_     0.08112      b±_  0.002704&, 

9   ~  1  +vt  b  ~  1  +0.00204*'30~  1  +0.00204*      S*     (       ) 


COMPRESSIBLE     OH     AElUfOKM     FLUIDS. 


275 


Whenever  the  elasticity  ol  the  air  is  expressed  by  the  press- 
ure p=  14.75  lbs.  on  the  square  inch,  instead  of  the  barometric 
height  b,  the  density  for  any  other  tension  p1  will  be 


jt?,_     0.08112        p,  0.0055;?, 

1  +0.00204*'14.75~  1  +0.00204* 


lbs.    (163) 


ri     1+vp  p 

The  density  of  steam  is  five  eighths  of  the  density  of  atmos- 
pheric air  for  the  same  temperature  and  tension.  Therefore 
we  have,  for  steam, 


0.00344/? , 
1+0.00204* 


lbs. 


417.  Prop.  To  determine  the  height  corresponding  to  a  given 
density  of  the  atmosphere,  and,  conversely,  the  density  in  terms 
of  the  height. 

Since  the  density  of  the  atmosphere  at  the  surface  of  the 
earth  is  due  to  the  pressure  of  the  superincumbent  portions  o 
it,  the  density  must  decrease  as  the  height  increases. 

Let  HOBA  be  a  vertical  column  of  air  whose 
base  AB  is  one  square  foot.  Conceive  it  divided 
into  portions  of  equal  weights  w,  and  heights  x  0 ,  x , , 
x2,  &c,  beginning  at  AB,  so  small  that  the  den- 
eity  of  each  may  be  regarded  as  uniform.  Then 
x0+xl+x2,  &c.  .  .  .  xn—^.x=h  is  the  height  of 
the  nth  stratum.  Let  p0  be  the  tension  of  the 
lowest  stratum,  and  pn  that  of  the  n\\\,  p0  and  pn 
their  densities  respectively.  Then  the  weight  of 
the  nth  stratum  is 


A 


B 


xnp  npn 

P  0 


(a) 


Po    Pn 

But  the  tension  of  the  (n— l)th  stratum  is  p„-\~ip,  and  the 
height,  therefore, 

Po       w 


2*n-;  = 


Pb'Pn+w' 


276  HYDROSTATICS. 

in  like  manner,  ar„_9=='-2-. — rr— , 

p0  p„+2w 


^Po       w      =  p,  w 
°"    X"~n     P0'Pn+nw     p0>0' 
since  pn+nw=p  Q. 

Now,  in  the  exponential  series, 

when  y  is  very  small,  we  have 

e^l+y, 
or  y=yle=l(l+y)=2.3025SL.(l+y),  (b) 

in  which  7  denotes  the  Naperian  logarithm,  and  L  the  common 

logarithm  of  a  number. 

w 
If  in  (b),  for  y,  we  put  — ,  we  get 

— =7.(i  +— )=Z.  M- )=/.(/vfu>)-Z..p.. 

.*.  from  (a),  we  have 

P  0 

Now,  substituting  j9n+w  successively  for  J9„,  we  obtain 
xn    =—\l(pn+w)-lpn\, 

Po 

xn_l=^{l(pn+2w)-l(pn+w)\, 

r  0 

r  n 


a:0=xn_n=^{/.(7>„.+ww)-Z.(pn+(n-l)u?)}. 

r  0 

By  taking  the  sum  of  these  equations,  the  terms  in  the 
brackets  will  all  cancel,  except  h(p„+nw)  =  lpt  in  the  last 
and  lp„  in  the  first,  and  we  shall  have 

Po      r  Po    P» 


COMPRESSIBLE     OR     AERIPORM     FLUIDS.  27? 

To  find  pn  when  h  is  given,  we  have 

P»  Po 

or  ^=e*°. 

P* 

_p1h 

Hence  p,=p0.e  Po ,  (165) 

where  e=2.71828,  the  base  of  the  Naperian  system  of  loga- 
rithms. 

418.  Schol.  Formula  (164)  may  be  adapted  to  the  determ- 
ination of  heights  by  the  barometer.  To  this  end  let  b0  and  b„ 
be  the  heights  of  the  barometrical  columns  at  the  lower  and 

v       b 
upper  stations  respectively.     Then,  since  — °— ir* 

/*=2.30258^.l|5,  (a) 

The  value  of  —  may  be  determined  from  the  consideration 
Po        J 
thatp0  expresses  the  weight  or  pressure  of  a  column  of  the  at- 

P 
mosphere  on  a  unit  of  surface  as  one  square  foot,  and  — -  must 

Po 

express  the  height  of  this  column  on  the  supposition  that  its 
density  is  uniform.  Now  a  cubic  foot  of  air  at  32°  weighs 
0.08112  lbs.,  and  a  cubic  foot  of  water  at  the  same  temperature 
weighs  62.37917  lbs.,  and  therefore  the  specific  gravity  of 

62.37917 

water  referred  to  air  is      '  —769.     Hence  the  height  of 

U.Uo  1  I  & 

a  homogeneous  atmosphere  at  a  temperature  of  32°  is  34x769 
=26146  feet. 

.-.  2.30258— =2.30258X26146=60204  feet. 

Po 

It  is  here  assumed,  however,  that  the  lower  station  is  at  or 
near  the  level  of  the  sea,  and  no  account  is  taken  of  the  varia- 
tion of  gravity  at  different  elevations.  From  numerous  obser- 
vations made  at  different  elevations  above  the  sea,  and  at  known 
differences  of  height,  this  coefficient  is  found  to  be  60345  fee/ 


278  il  Y  D  a  O  S  T  A  T     US. 

at  a  temperature  of  32°.  But  the  actual  temperature  of  the  an 
at  both  the  lower  and  upper  stations  will,  in  general,  offer  from 
the  standard  temperature  of  the  formula,  and,  since  the  density 
of  air  varies  uniformly  with  the  temperature,  we  may  use  the 
mean  of  the  temperatures  of  the  air  at  the  two  stations.  Let 
t0  and  tni)e  the  indications  of  Fahrenheit's  thermometer;  then 
the  mean  temperature  will  be  l(t0+t,),  and  the  deviation  t  from 
the  standard  will  be 

t=l(t0+t„) -32. 

The  expansion  of  dry  air  is  0.00204  for  a  change  of  1°;  but 
when  the  atmosphere  contains  vapor,  it  is  found,  by  comparing 
the  rates  of  expansion  of  vapor  and  of  dry  air,  and  assuming  a 
certain  mean  humidity  for  the  air,  that  the  rate  is  expressed  by 
0.00222.  Incorporating  this  correction  in  (a),  and  using  the 
coefficient  determined  by  observation,  we  have 

/*=60345(1  +. 002220  L.t^.  (166) 

It  is  further  obvious  that  a  change  in  the  length  of  the  mer- 
curial column  will  be  produced  by  a  change  of  temperature  of 
the  mercury.  Let  r0  and  t„  be  the  temperatures  of  the  mer- 
cury, as  shown  by  a  thermometer  attached  to  the  cistern  of  the 
barometer;  then,  since  mercury  expands  at  the  rate  of  0.0001 
for  each  degree, 

b=b'lt(l+. 0001)  {r0-rn), 

where  b'n  is  the  observed  height  of  the  barometer  at  the  upper 
station.  Using  this  value  of  bn,  the  difference  of  elevation  be- 
tween two  stations  or  the  height  of  a  mountain  may  be  determ- 
ined with  considerable  accuracy. 

In  the  determination  of  the  constant  coefficient,  the  variation 
of  gravity  at  different  elevations  is  allowed  for  in  the  assump- 
tion that  this  coefficient  is  that  which  belongs  to  the  mean 
height  above  the  sea  at  which  observations  are  usually  made, 
and  to  the  latitude  of  45°.  When  the  latitude  differs  from  this, 
it  will  be  necessary  to  multiply  the  result  by 

(1 +.002837  cos.  2t/>), 
i/>  being  the  latitude  at  which  the  observations  are  made. 


COMPRESSIBLE     OR     AERIFORM     FLUIDS.  279 


419.    EXAMPLES. 

Ex.  1.  A  cylindrical  tube  40  inches  long  is  half  rilled  with 
mercury,  and  then  inverted  in  a  vessel  of  mercury.  How 
high  will  the  mercury  stand  in  the  tube,  the  pressure  of  the  ex- 
ternal air  being  equivalent  to  30  inches  ? 

Let  /  be  the  length  of  the  tube,  a  the  length  of  the  portion 
occupied  by  the  air  before  it  was  inverted,  h  the  height  of  mer- 
cury due  to  the  pressure  of  the  external  air,  x  the  height  of  the 
mercury  after  the  tube  is  inverted,  and  h'  the  column  of  mer- 
cury equivalent  to  the  tension  of  the  air  in  the  tube. 

Then  (157),  ha=h'(l-x). 

But  h=h'+x  or  h'=h—x. 

.*.  ha=(h—x)  (l—x), 
whence  we  get  x=\{l-\-h)±\  V(l— xy+4ah. 

Using  the  data  of  the  question,  /=40,  a— 20,  and  h=S0,  we  get 

a,-=10  or  60. 
The  first  value  is  that  which  pertains  to  the  specific  question. 

Ex.  2,  A  tube  30  inches  long,  closed  at  one  end  and  open  at 
the  other,  was  caused  to  descend  in  the  sea  with  the  open  end 
downward  until  the  inclosed  air  occupied  only  one  inch  of  the 
tube.     How  far  did  it  descend  ? 

Ex.  3.  A  spherical  air-bubble  having  risen  from  a  depth  of 
1000  feet  in  water,  was  one  inch  in  diameter  when  it  reached 
the  surface.     What  was  its  diameter  at  the  bottom  ? 

Ex.  4.  Required  the  equation  of  the  curve  described  by  the 
extremities  of  a  horizontal  diameter  of  the  air-bubble  of  Ex.  3, 
supposing  its  center  to  move  in  a  vertical  line. 

Ex.  5.  What  number  of  degrees  must  a  given  volume  of  air 
be  heated  to  double  its  elasticity? 

Ex.  6.  The  following  barometrical  observations  were  made 
at  the  White  Rocks,  on  the  bank  of  the  Connecticut  River 
some  two  miles  below  the  city  of  Middletown  : 

Pyrometer.  Det.  Ther.  At  Ther. 

At  the  base,        60  =  30.09in.       t0=83.0       r0  =  84.5 
On  the  summit,  b'„  =29.65  in.       tn  =85.0       t„  =83.5 


280 


H  Y  D  11  O  S  T  A  T  I  C  S. 


Hence 
Also, 


t—%(t0+L)—32°=52°  and  1 +.00222/  =  1.1 1544. 

t0-t„=1°  and  6n=6:,(l+.OOOlXl)  =  29.653  in. 

,      „  T   30.09 

.'.  //  =  60345 X  1.11 544. XL 


29.653 


30.09  log.   1.4784222 

29.653  a.c.  "      8.5279314 


0.0063536  log.  7.8030199 
1.11544  "  0.0474462 
60345.  "     4.7800413 

427.67  feet   "     2.6311074. 

Ex.  7.  The  following  observations  were  made  by  Humboldt 
at  the  Mountain  of  Quindiu,  New  Grenada,  in  lat.  5° : 

Barometer. 

At  the  level  of  the  Pacific,  30.036  in.  IT 

On  the  summit,  20.0713  6^ 

Required  the  height  of  the  mountain. 

Ans.  11500  feel  neaiiv. 


Det.  Ther. 

n"1r0.54 


At.  Ther. 

77°.54 
75       08°.00 


£ 


HYDRODYNAMICS. 


420.  Prop.  The  velocity  of  a  fluid  in  a  tube  of  variable  di* 
ameter,  kept  constantly  full,  is  in  different  transverse  sections  in- 
versely as  the  areas  of  the  sections. 

Since  the  tube  is  supposed  constantly  full,  and  the  fluid  in- 
compressible, the  same  quantity  of  fluid  must  pass  through 
every  section  in  a  unit  of  time.  But  admitting  the  fluid  to  have 
the  same  velocity  in  every  part  of  the  same  section,  the  quan- 
tity which  flows  through  any  section  in  a  unit  of  time  will  be 
the  product  of  the  area  of  the  section  by  the  velocity.  If, 
therefore,  k  and  k'  be  the  areas  of  any  two  sections,  and  v  and 
v'  the  velocities  at  each  respectively, 

kv=k'v', 
and  v  :  v'=k'  :  k.  (167) 

In  reality,  the  velocity  is  diminished  by  the  sides  of  the  tube, 
and  is  therefore  in  any  section  least  near  the  sides  of  the  tube 
and  greatest  near  the  central  portions. 

421.  Prop.  The  velocity  with  which  a  fluid  issues  from  a  small 
orifice  in  the  bottom  of  a  vessel  kept  constantly  full,  is  equal  to 
that  acquired  by  a  body  falling  freely  through  the  height  of  the 
fluid  above  the  orifice. 

Let  .£F  represent  a  very  small  orifice  in  ^  lm  b 
the  bottom  of  the  vessel  ABCD,  filled  with  a 
fluid  to  the  level  of  AB,  GF  a  stratum  whose 
thickness  FH=/i'  is  indefinitely  small,  and 
FM=A  the  whole  height  of  the  column  ver- 
tically above  the  orifice. 

If  now  the  stratum  GF  fall  by  its  own 
weight  through  HF=/j',  the  velocity  will  be 


T) 


G- 


E  F 


282  HYDRODYNAMICS. 


v=  V2gh'. 
But  if  the  stratum  be  urged  by  its  own  weight  and  the  weight 
of  the  column  LH,  calling  the  force  in  this  case  g',  the  velocity 
v  will  be 


v'=  V2g'h'. 

But  the  forces  g  and  g'  are  as  the  weights  of  the  columns 

GF  and  LF,  or  as  their  heights  h'  and  h.     Hence 

g'     h  gh 

— =—  or  £•'  =  —. 
g     h'      8     h1 

Substituting  this  value  of  g1  in  that  of  v',  we  have 

v'=V2gh,  (168) 

which  is  the  velocity  of  a  body  falling  freely  through  the 
height  h. 

Cor.  If  the  orifice  be  made  in  the  side  of  the  vessel,  and  a 
tube  be  inserted  so  as  to  direct  the  current  obliquely,  horizon- 
tally, or  vertically  upward,  the  velocity  of  efflux  will  be  the 
same,  since  the  pressure  of  fluids  is  the  same  in  every  direc- 
tion. 

In  the  first  case,  its  path  will  be  a  parabola  whose  equation 
is  (G4). 

In  the  second  case,  the  angle  of  elevation  a=0,  which  re- 
duces the  equation  to 

x>=4hy,       » 
the  equation  of  a  parabola  whose  axis  is  vertical  and  origin  of 
co-ordinates  at  the  vertex  or  orifice. 

In  the  last  case,  if  all  obstructions  are  removed,  the  fluid  will 
rise  to  the  height  of  the  surface  of  the  fluid  in  the  vessel. 

422.  Prop.  To  determine  the  horizontal  distance  to  which  a 
fluid  will  spout  from  an  orifice  in  the  vertical  side  of  a  vessel. 

Let  the  vessel  ABCD  be  filled  to  the  level  AB.  If  the  fluid 
issue  horizontally  from  the  orifice  O,  the  equation  of  its  path  is 

x* =4  hy, 
in  which  /j  — OB,  the  height  of  the  fluid  above  the  orifice,  is  the 
impetus  or  height  due  to  the  velocity. 

To  determine  the  range,  let  y—OC=a.     Hence 


DISCHARGE     1       OM     SMALL     ORIFICES. 


283 


x=CP=2  y/hy=2  Vha=2.0M, 
or  the  horizontal  distance  is 
equal  to  twice  the  ordinate  at 
the  orifice,  in  a  semicircle  whose 
diameter  is  the  vertical  distance 
from  the  surface  of  the  fluid  to 
the  horizontal  plane. 

Cor.  When  the  orifice  is  at 
the  middle  of  BC,  the  range  is  a 
maximum,  and  at  equal  dis- 
tances above  and  below  the  mid- 
dle the  range  will  be  the  same. 

423.  Prop.  To  find  the  quantity  of  fluid  discharged  in  a 
given  time  from  a  small  orifice  in  the  bottom  of  a  vessel  when 
the  fluid  is  maintained  at  the  same  constant  height. 

Let  h  be  the  constant  height  of  the  fluid  in  the  vessel,  t  the 
given  time,  k  the  area  of  the  orifice,  and  Q,  the  quantity. 

Then  Q,=kvt=ktV2gh.  (169) 

If  the  orifice  be  circular,  and  r  its  radius,  then  k=nr*,  and 
Q=n  V~2g.t?'W7i, 
=25.195.trQ  Vh. 
The  time  being  in  seconds,  and  r  and  h  in  feet,  Q,  will  be  in 
cubic  feet.     If  the  weight  W  be  required, 

W=62.5Q. 
Cor.'  1.  The  time  required  for  the  efflux  of  a  given  quantity  is 

nr*  V2gh 
Cor,  2.  Since  t  and  g  are  constant  in  (169), 

Hence  the  quantity  discharged  in  the  same  time  from  orifices 
differing  in  size  and  distance  from  the  surface,  varies  as  thf 
size  of  the  orifice  and  square  root  of  its  depth. 

If  the  orifices  are  the  same,  QxVh.  In  order,  therefore, 
(hat  the  discharge  from  one  orifice  maybe  twice  that  from  an- 
other, its  depth  must  be  four  times  as  great. 


284  HYDRODYNAMICS. 

424.  Prop.  To  determine  the  time  in  which  a  cylindrical  ves- 
sel filled  with  water  will  empty  itself  by  a  small  orifice  in  its 
base. 

The  velocity  of  efflux  at  any  instant  is  v=  V2gh,  h  being  the 
height  of  the  fluid  in  the  vessel.  But  since  the  velocities  v  and 
v',  at  the  orifice  and  at  any  transverse  section  of  the  vessel,  are 
inversely  as  their  areas,  k  and  k'  (167), 

h       fa     _ 

v'=y,v—  FT  Vr2^r/t  oc  Vh. 
K        k 

The  surface  of  the  fluid,  therefore,  in  the  vessel  descends  with 
a  velocity  proportional  to  the  square  root  of  the  space  over 
which  it  must  pass,  and  the  circumstances  of  its  motion  are 
precisely  the  same  as  those  of  a  body  projected  vertically  up- 
ward. But  the  velocity  of  a  body  projected  upward  is  such 
that  if  it  continued  uniformly  it  would  move  over  twice  the 
space  through  which  it  must  move  in  the  same  time  to  acquire 
or  expend  that  velocity.  Hence  the  time  of  descent  of  the 
surface  of  the  fluid  to  the  base  is  twice  that  required  for  the 
descent  of  the  same  superficial  stratum  when  the  vessel  is  kept 
full.  In  the  latter  case,  if  Q,  be  the  whole  quantity  of  water 
in  the  vessel  (169), 

.=-3= 

kV2gh 

Therefore,  in  the  former, 

*=-?S=.  (170) 

kV2gh 

If  r'  be  the  radius  of  a  tranverse  section  of  the  vessel,  r  thai 
of  the  orifice, 

Q=7rr'2A  and  i=nr*. 

.\  t= — =Vh. 
rW2g 

425.  Schol.  The  preceding  deductions  are  founded  on  the 
hypothesis  that  the  fluid  particles  descend  in  straight  lines  to 
the  orifice,  and  ah  issut'  with  a  velocity  due  to  the  height  of 
the  fluid  surface.     Experiment  shows  that  this  is  true  only  of 


DISCHARGE     FROM     SMALL     ORIFICES.  285 

those  vertically  above  the  center  of  the  orifice,  that  those  sit- 
uated about  the  central  line  of  particles,  take  a  curvilinear  course 
as  they  approach  the  orifice,  being  turned  inward  toward  this 
line  or  spirally  around  it,  and  this  deviation  from  a  vertical 
rectilinear  path  is  the  greater  the  more  remote  they  are  from 
the  central  line.  This  deviation  will  necessarily  occasion  a 
diminution  of  vertical  velocity,  and  therefore  a  diminution  of  to- 
tal discharge.  The  oblique  direction  of  the  exterior  particles 
gives  to  the  vein  of  issuing  fluid,  when  the  orifice  is  circular, 
the  form  nearly  of  a  conic  frustum,  whose  larger  base  is  the 
area  of  the  orifice.  This  diminution  of  a  section  of  the  issuing 
fluid  is  called  the  contraction  of  the  vein,  and  the  vein  itself, 
from  the  orifice  to  the  smallest  section,  the  vena  contracta,  or 
contracted  vein. 

The  results  of  most  experiments  agree  in  making  the  length 
of  the  contracted  vein,  when  the  orifice  is  circular  and  hori- 
zontal, equal  to  the  radius  of  the  orifice.  There  is  a  greater 
discrepancy  in  the  results  of  experiments  for  determining  the 
ratio  of  the  diameters  of  the  two  ends  of  the  contracted  vein. 
When  water  flows  through  orifices  in  thin  plates  it  is  found  to 
be  about  0.8.  The  ratio  of  the  areas  of  the  two  ends,  called 
the  coefficient  of  contraction,  is  therefore  0.64. 

The  actual  discharge  is  found  by  experiment  to  differ  slight- 
ly from  the  theoretical  for  other  causes.  The  ratio  of  the 
former  to  the  latter  is  found  to  be  about  0.97,  and  is  called  the 
coefficient  of  velocity.  The  product  of  the  coefficients  of  ve- 
locity and  contraction,  called  the  coefficient  of  efflux,  is  0.62. 

Hence,  for  the  actual  velocity  of  discharge  through  orifices 
in  thin  plates,  we  have 

vl  =0.62^=0.62  V2gh, 
and  for  the  quantity  in  a  unit  of  time  (169), 

Q=kv1=0.G27iV~2gh. 
When  the  orifice  or  pipe  through  which  the  discharge  is  made 
has  the  length  and  form  of  the  vena  contracta,  the  velocity 
will  be  v ,  =0.97u=0.97  V2gn, 

and  the  quantity       Q=kvl=0.97kV2gh, 
h  being  the  smallest  section  of  the  contracted  veir. 


280 


II  Y  D  R  O  D  i'  N  A  M  I  C  S. 


By  the  use  of  cylindrical  or  conical  adjutages,  the  quantitv 
of  the  discharge  is  increased.  More  seems  to  be  gained  by 
the  adhesion  of  the  fluid  particles  to  the  sides  of  the  tube,  in 
preventing  the  contraction  of  the  vein,  than  is  lost  by  the  fric- 
tion. The  discharge  is  greater  when  the  adjutage  is  conical 
and  the  larger  end  is  the  discharging  orifice. 

426.  Prop.  To  determine  the  quantity  of  water  which  will 
flow  from  a  rectangular  aperture. 

1°.  When  one  side  of  the  aperture  coincides  with  the  sur- 
face of  the  water. 

Let  h  be  its  height  and  b  its  breadth,  and 
conceive  the  aperture  to  be  divided  horizon- 
tally into  a  very  large  number  n  of  equal  di- 
visions so  narrow  that  the  velocity  of  the 
fluid  in  every  part  of  each  may  be  regarded 
as  the  same.  The  larger  n  is,  the  more  near- 
ly will  the  hypothesis  be  satisfied. 

The  depth  of  these  successive  divisions  be- 


low the  surface  will  be 


h    2h    3h 


n     n      n 
The  velocities  in  each  will  be 


,  &c. 


/hi    2h      /    3h   0 


and  since  the  areas  of  these  divisions  are  each  b-,  the  quanti* 

n 

ties  discharged  by  each  in  a  unit  of  time  will  be  (169) 
bh    f~h    bh     I    2h   bh     J    3h    „ 

tv^  tv  2=v  tv  ^t  &c- 

and  their  sum,  or  the  whole  quantity  Q,  discharged,  will  be 


But 


^     bhV2gh    i      i      i 

Q= -^=(l2+22  +  32+,  &c.  . 

nvn 

l2+22+3^+,  &c.  . 


I     /r 


4+i 


1  +  1 


In2. 


DISCHARGE     FROM     RECTILINEAR     APERTURES.    28T 


bh 


...  Q=|6/t  V2gh=°~b  V2gh\ 
Or,  if  15=7^  be  the  mean  velocity, 

2°.  When  the  upper  side  of  the  rectangular  aperture  is  not 
coincident  with  the  surface  of  the  fluid. 

Let  its  depth  EL=/ji5  and  the  distance  EG=A,  as  before. 
Then  the  quantity  which  issues  from  the  aperture  LH  will  be 
equal  to  the  quantity  which  would  issue  from  EH,  diminished 
by  the  quantity  which  would  issue  from  EK,  or 


q=lbV2gh3-^V2ghl  =  ^bV2g(h2-h21); 
and  if  Q=b(h— A,)u, 


v=%V2e. 


h—h1 


427.  Prop.    To  determine  the  quantity  of  water  which  will 
flowf?'om  a  triangular  aperture. 

1°.  When  the  vertex  of  the  triangle  is  in  the  surface. 

Let  the  base  GH=&,  and  the  height  HF=/j,        v        ^ 
and  let  the  triangle  be  divided  into  n  parts,  as 
before,  of  equal  but  very  small  heights.     The 

altitudes  of  these  elements  are  each  -,  and  the 


II 


lengths,  beginning  at  the  vertex  of  the  trian 

i  b  2b   3b 

ffle,  are  -,  — ,  — ,  otc.     1  heir  areas,  regard 
°    '  n    n     n 

ing  them  as  parallelograms,  which  we  may  do,  since  n  is  in 

definitely  large,  are 

bh   2b  h   3b  h    c 
-.-,  — .-,  — .-,  &c. 
nn    n  n    n  n 

Hence  the  discharges  through  each  are 

bh    /~~h   2bh    /~~2k    Sbh    /~~3h    . 

and  the  whole  discharge  will  be 


288 


HYDRODYNAMICS. 


Q  = 


bhy/2gh    a      §      g 

= —-(l2+22+32+,  &c. 

n*  Vn 

3,  .  -. 

bhV2srh/n2 


n2) 


Also, 


hV2gh/7i-     \ 
nWn  Vf+1' 


fMV2gh=%bV2gh3. 


2°.  When  the  base  of  the  triangle  is  in  the  surface. 
In  this  case  the  quantity  which  will  flow  through  EFG  wil, 
equal  the  discharge  through  EFGH  minus  that  through  FGH, 

or  Q/=fbhV2gh-^bhV2gh=T%bhV2gli, 

and  v=^jV2gh. 

Cor.  If  the  aperture  be  a  trapezoid,  as 
EFGH,  whose  upper  base  EF=&j  lies  in 
the  surface,  lower  base  HG=fe2,  and  alti- 
tude KH=/i,  we  may  find  the  discharge  by 
regarding  the  aperture  as  made  up  of  the 
rectangle  KHGM,  and  the  two  triangles 
EKH  and  MGF.     Hence 

Q^=^bJiV'2^+T%(b1-b2)V'2gTi==T2j{2b1+3b2)hV'2gh. 
Also,  through  'the  triangle  HOG,  whose  base  HG=&2,  and 
depth  of  vertex  hO—hl,  the  quantity  of  the  discharge  will  be 

Q=tW'i  V2^7-T2J(2&1+3&2)Av/2i:A. 

In  a  similar  manner,  rectilinear  orifices  of  other  forms  may 
be  divided  into  triangles,  trapeziums,  &c,  and  the  discharge 
determined. 

428.  Prop.  To  determine  the  velocity  with  which  an  elastic 
fluid  will  issue  from  a  small  orifice  into  an  unlimited  void  when 
urged  by  its  own  weight. 

By  reference  to  Art.  421,  it  will  be  seen  that  the  result  there 
obtained  is  independent  of  the  density  of  the  fluid,  and  will 
therefore  be  true  of  all  fluids,  whatever  be  their  density.  For 
the  forces  g  and  g'  are  as  the  weights  of  the  stratum  at  the 
orifice,  and  of  the  column  vertically  above  it;  that  is, 
g  :  g'—w  :  io'=gh  :  gh'=h  :  h'. 


MOTION     OP     FLU    iDS     IN     LONG     PIPES.  289 


Consequently,  v--  V2gh' 

expresses  the  velocity  with  which  every  fluid  of  uniform  dens 

ity  will  issue  from  a  small  orifice. 

Cor.  The  velocity  with  which  air  will  rush  into  a  vacuum 
is  that  which  a  heavy  body  would  acquire  in  falling  from  the 
height  of  a  homogeneous  atmosphere.  The  height  of  a  homo- 
geneous atmosphere  {Art.  418),  when  the  temperature  is  32° 
and  the  barometer  stands  at  30  in.,  is  /i'  =  26146  feet,  which 
gives  u=1297  feet.  But  for  a  temperature  of  60°  the  height 
A,  will  be  (159) 

h1=h'(l+vt), 
in  which  *=60°  —  32°  =  28°,  and  the  value  of  v  for  a  mean  state 
of  humidity  of  the  air  will  be  0.00222  instead  of  0.002039 
(Art  412). 

.'.  v=  V2ghl=  V2gh'(l  +0.00222X28) 


=  V2X32|X2614GX(1+0.00222X28)  =  1336  feet 

MOTION    OF    FLUIDS    IN    LONG    PIPES. 

429.  The  subject  of  the  conveyance  of  water  in  pipes  is  one 
of  considerable  practical  importance.  But  the  motion  of  fluids 
in  long  pipes  is  so  much  affected  by  adhesion  and  friction  in 
the  interior,  by  the  resistance  occasioned  by  curvature,  and  by 
the  disengagement  of  air,  which  remaining  stationary  when 
the  pipes  are  laid  along  a  level  surface,  or  collecting  in  the 
higher  portions  of  them,  where  they  are  curved,  opposes  the 
flow  of  the  fluid,  that  theoretical  results  are  of  little  practical 
value ;  besides,  investigations  conducted  on  hypotheses  in- 
volving all  the  causes  which  affect  the  motion  of  the  fluid  are 
too  difficult  for  an  elementary  work. 

The  experiments  of  Bossut,  in  1779,  are  those  which  are 
usually  relied  on  for  information  on  this  subject.  Water  was  al- 
lowed to  flow  through  pipes  whose  diameters  were  li  in.  and 
2  in.,  and  lengths  from  30  to  180  feet.  They  were  chiefly  of 
tin,  and  inserted  in  the  side  of  a  reservoir  filled  to  a  constant 
height,  either  one  foot  or  two  feet  above  the  center  of  the  pipe 
The  following  principles  were  established : 

1°.  The  discharges  in  given  times  with  pipes  of  the  same 

T 


kOO 


HYDRODYNAMICS. 


length,  and  with  the  same  head  of  water,  are  proportional  to 
the  squares  of  the  diameters. 

2°.  For  pipes  of  different  lengths  and  of  the  same  diameter, 
the  discharge  was  inversely  proportional  to  the  square  roots 
of  the  lengths. 

To  determine  the  supply  which  may  be  expected  from  a 
pipe  of  given  dimensions,  it  was  found  that  a  pipe  30  feet  long 
and  li  inch  in  diameter  would  discharge  at  its  extremity  about 
one  half  of  that  which  would  issue  from  a  simple  orifice,  or 
short  pipe  of  the  same  diameter. 

Couplet,  in  1730,  found  that  a  pipe  of  stone  or  iron,  600  fath- 
oms in  length  and  12  inches  in  diameter,  with  a  head  of  12  feet 
of  water,  discharged  TVth,  and  a  pipe  of  the  same  diameter  and 
2340  fathoms  in  length,  with  a  head  of  20  feet,  discharged  TVh 
of  that  which  would  have  been  obtained  from  a  simple  orifice. 

To  determine  the  quantity  which  flows  through  a  section  of 
a  natural  stream,  it  is  usual  to  measure  the  breadth  and  the 
depth  at  different  points  of  a  transverse  section,  and  find  the 
area  of  the  section.  The  velocity  at  various  points  of  the  sec- 
tion is  measured  by  the  hydraulic  quadrant,  or  rheometer  of 
some  kind.  The  mean  velocity,  multiplied  by  the  area,  gives 
the  quantity  which  flows  through  the  section  in  a  second. 


430.  Prop.  To  determine  generally  the  velocity  with  which  a 
fluid  will  issue  from  an  orifice  of  any  size  in  the  bottom  of  a 
cylindrical  vessel. 

A  We  shall  adopt  the  usual  hypothesis  of  a 

division  of  the  fluid  into  thin  lamina?,  and  thai 
in  their  descent  their  parallelism  is  preserved. 
■P'  Let  the  distance  of  the  lamina  pp'qq'  from 
the  surface  A  be  x,  and  K  the  area  of  its 
surface.  Then,  if  dx  be  its  thickness  and  p 
its  density,  pKdx  will  be  its  mass,  and  gpKdx 
its  weight,  or  the  force  with  which  it  wili 
descend  if  free.  Now  the  resistance  with 
which  it  meets  in  its  descent  is  the  difference 
of  pressures  on  its  lower  and  upper  surfaces. 
Let/>  represent  the  pressure  on  a  unit  of  tht 


f 

c 

D  B  E 


GENERAL     METHODS.  291 

upper  surface  by  the  water  above  it ;  then  p+dp  will  be  the  up- 
ward pressure  on  a  unit  of  the  lower  surface,  or  the  resistance 
of  the  water  below  it,  and  the  difference  —dp,  multiplied  by 
K,  will  be  the  resistance  experienced  by  the  whole  lamina. 
Hence  the  moving  force  will  be 

gpK.dx-K.dp, 
and  the  acceleration,  [VI.]  and  [IX.], 

_gpKdx— Kdp    gpdx  —  dp     d2x 
0=       ~pKdx       =     pdx     =~dJ'  (a) 

Let  v  be  the  velocity  of  discharge  at  B,  k  the  area  of  a  sec- 
tion of  the  issuing  fluid,  and  dt  the  element  of  time  in  which 
the  surface  K  descends  a  distance  equal  to  dx.  Then  (167) 
we  have 

„  ,       7    7        dx    kv 

Kdx=kvdt  or  -jr—r?--  (o) 

dt     K  v  ' 

By  differentiation,  regarding  dt  as  constant,  we  get 

d2x  __kdv 

~dT=Kdt 

kdv  _gpdx  —  ap 

Kdt         pdx 

_        ,      pkdv  dx 
gPdx-dp=-w.Tt. 

k* 
Hence  (b)  gpdx—dp—  p^.vdo. 

Integrating,  gpx—p=p^.—+C. 

Let  now  a;=A=AB  and  k=K  ;  then 
gph-p=p-+C ; 
and  subtracting  the  preceding  from  this,  we  have 

g{h-x)=-{l--), 

-v7^.        („„ 

1_Ka 


292  HYDRODYNAMICS. 


If  k  be  very  small,  v=  V2g(h—x)  nearly;  and  if  the  vessel  be 
kept  constantly  full,  x=0  and  v=  V2gh  (168). 

If  k=K,  v=  ac,  or  the  velocity  must  be  infinite.  From  which 
we  infer  that  a  section  of  the  issuing  fluid  can  never  be  equal 
to  a  section  of  the  vessel.  If  a  cylindrical  tube  be  vertical  and 
filled  with  a  fluid,  the  portion  of  the  fluid  at  the  lower  extremi- 
ty, being  urged  by  the  pressure  of  all  above  it,  will  neces- 
sarily have  a  greater  velocity  than  those  portions  which  are 
higher,  and  therefore  (167)  a  section  of  the  issuing  fluid  is  nec- 
essarily less  than  a  section  of  the  tube. 

431.  Prop.  To  determine  the  time  in  which  a  cylindrical  ves 
sel  will  exhaust  itself  by  a  small  orifice  in  the  base. 

Since  k  is  very  small,  we  have  v—  V2g(h—x)  for  the  veloc- 
ity at  the  end  of  the  time  t  when  the  surface  of  the  fluid  has 
descended  through  the  space  x.  Let  this  velocity  be  sup- 
posed constant  during  the  indefinitely  small  time  dt.  Then 
the  quantity  discharged  in  this  time  will  be 

tfQ, = kdt  V2g(h—x)  =  Kdx, 

dx  being  the  descent  of  the  surface  in  the  time  dt. 

K  _i 

Hence  dt— — =.(h—x)  -dx. 

kV2g 

2K  i 

Integrating,  t= —.(A— x)2+C. 

kV2g 

2K      - 
If  x=0,  *=0,  and  C= — =:Vh. 
kV2g 

2K    ,    - 


.-.  t= — =(Vh-Vh-x),  (172) 

kV2g 

and  when  x=h,  or  the  whole  is  discharged, 

kV2g     kV2gh    kV2gh 

The  time  is  therefore  twice  as  great  as  that  which  is  required 
to  discharge  the  same  quantity  when  the  vessel  is  kept  con- 
stantly full. 


EXAMPLES.  293 

432.  Pitop.  To  determine  the  quantity  which  will  issue  from 
an  aperture  of  any  size  and  form  in  the  side  of  a  prismatic 
vessel. 

Let  the  Fig.,  Art.  430,  represent  one  face  of  the  vessel,  and 
DCE  the  aperture. 

The  element  of  the  area  of  the  orifice  will  now  be  ydx. 

.-.  dQ=tydxV2g(h1+x), 
in  which  hl=AC,  x=Cm,  and  y=mn. 

This,  integrated  between  the  limits  x~0,  and  x=h—h  l—CB 
will  give  the  quantity  discharged  in  the  time  t. 

If  the  aperture  be  rectangular,  y—a.  constant=&.     Then 

dQ=bt  V2g(h1+z)  dx, 

and  Cl=^btV:2g(h1+xf+C, 

and,  between  the  limits  above  named, 

Q=f&*V%r(Af-4).  (1^4) 

If  the  orifice  extend  to  the  top  of  the  vessel,  then  h  ,  =0,  and 

.       3  

Q=f6f  V2g.h2=%bht  V2gh, 

v=m=%V2gh, 

which  is  the  velocity  due  to  the  depth  of.  the  center  of  pressure 
of  the  aperture  below  the  surface  of  the  fluid. 

In  the  foregoing  the  vessel  is  supposed  to  be  kept  constant- 
ly full,  or  the  surface  of  the  fluid  to  be  very  large  compared 
with  the  aperture. 

433.    EXAMPLES. 

Ex.  1.  A  vessel,  formed  by  the  revolution  of  a  semi-cubical 
parabola  about  its  axis,  which  is  vertical,  is  filled  with  a  fluid 
till  the  radius  of  its  surface  is  equal  to  its  distance  from  the 
vertex  ;  to  find  the  time  in  which  the  fluid  will  be  discharged 
through  a  small  hole  at  the  vertex. 

The  equation  of  the  semi-cubical  parabola  is  ay1— x\ 

Hence  (167)  kvdt=  —  Kdx, 

Trx^dx 

or  k  V  2gxdt  =  —  ny^dx  = . 


294  HYDRODYNAMICS. 

.*.  at= ^.x~dx. 

kaV2g 

Taking  the  integral  between  the  limits  x=0  and  v=a  we 
have 

~  i     7:V2ai 

t== z=z.^a-  = =r. 

kaV2g  Ik  Vg 

Ex.  2.  To  find  the  time  in  which  a  paraboloid  of  revolution 
whose  altitude  is  h  and  parameter  p,  full  of  fluid,  will  empty 
itself  through  a  small  orifice  at  its  vertex,  its  axis  being  vertical. 

3 

2-rrph2 

Ans.  t= =. 

3kV2g 

Ex.  3.  A  conical  vessel,  the  radius  of  whose  base  is  r  and 
altitude  h,  is  filled  with  a  fluid.  Required  the  time  in  which 
the  surface  of  the  fluid  will  descend  through  half  its  altitude, 
the  orifice  being  at  the  vertex  and  the  axis  vertical. 

7rr2F(2"-l) 
Ans.  t— j . 

20kg2 

Ex.  4.  Find  the  times  in  which  a  fluid  contained  in  a  vessel, 
formed  by  the  revolution  of  the  curve  whose  equation  is  yi=asx 
about  the  axis  of  x,  will  descend  through  equal  distances  h  , 
supposing  a  small  orifice  at  the  vertex,  and  the  axis  vertical. 

3 

_     ,  -na'h 

Ans.  .hacri  t= — ^. 
kV2g 

Ex.  5.  The  horizontal  section  of  a  cylindrical  vessel  is  100 
square  inches,  its  altitude  is  36  inches,  and  it  has  an  orifice 
whose  section  is  one  tenth  of  a  square  inch.  In  what  time,  if 
filled  with  a  fluid,  will  it  empty  itself,  allowing  for  the  contrac 
tion  of  the  vein  ? 

Ans.  t=nm3G'.5. 


HYDROSTATIC  AND  HYDRAULIC  INSTRUMENTS 


h 
A- 

B- 
C4 


M 


D 


MARIOTTE'S    FLASK. 

433.  This  piece  of  apparatus  was  devised  to  illustrate  several 
successive  positions  of  equilibrium  of  air  and 
water,  and  variations  in  the  velocity  of  dis- 
charge. In  the  annexed  section  of  this  flask, 
A,  B,  C  are  apertures  so  small  that  when  open 
and  water  is  issuing  the  air  can  not  enter,  and 
the  reverse.  The  apertures  at  first  are  all 
closed,  the  vessel  is  filled  with  water,  and  the 
tube  DF  is  inserted  at  M  and  filled  with  water 
to  the  height  D. 

1°.  Let  the  aperture  B  be  opened. 

If  h  be  the  height  of  a  column  of  water  equiva- 
lent to  one  atmosphere,  and  x  represent,  in  every 
case,  the  difference  of  level  between  the  surface 
of  the  water  in  the  tube  and  the  open  orifice,  then  the  pressure 
of  the  air  inward  at  B  on  a  unit  of  surface  will  be  h,  while  the 
pressure  outward  on  a  unit  of  surface  will  be  that  due  to  the 
pressure  of  the  air  on  the  surface  at  D,  together  with  the  press- 
ure of  the  column  of  water  DF,—x,  or  the  outward  pressure  at 
B  is  h+x.  Therefore  the  pressure  outward  is  greater  than 
the  pressure  inward  by  the  difference  h+x  —  h=x,  and  the 
water  will  issue  with  a  decreasing  velocity  due  to  x,  or 
v—  V2gx.  This  discharge  will  continue  till  x  vanishes  and 
the  water  in  the  tube  descends  to  E,  on  a  level  with  B,  when 
the  equilibrium  is  restored. 

2°.  Let  B  be  now  closed  and  A  be  opened. 

Now  the  pressure  downward  on  the  surface  of  the  water  in 
the  tube  at  E  is  h.  The  pressure  upward  on  the  same  surface 
is  that  due  to  the  press  are  of  the  air  at  A,  and  to  the  weight 


296     HYDROSTATIC    AND    HYDRAULIC    INSTRUMENTS. 

of  a  column  of  water  equal  in  height  to  AB=HE=:r,  or  the 
upward  pressure  on  E  is  h+x.  The  resultant  upward  press- 
ure on  E  is  therefore  h+x—h=x,  and  the  surface  of  the  water 
in  the  tube  will  rise  till  x  vanishes,  or  till  it  arrives  at  H,  when 
the  equilibrium  will  again  be  restored. 

In  the  mean  time,  however,  the  air  has  entered  at  A  and 
risen  to  the  upper  part  of  the  vessel,  where  it  occupies  the  space 
above  ho.  Since  an  equilibrium  now  exists,  the  expansive 
force  of  the  air  above  ho,  together  with  a  column  of  water 
equal  in  height  to  LH,  equilibrates  the  pressure  h  of  the  air  on 
the  water  in  the  tube  at  H.  If  h!  be  the  height  of  a  column 
of  water  equivalent  to  the  tension  of  the  air  above  ho,  ana 
y=LH  represent  the  difference  of  level  of  the  surfaces  of  the 
fluid  in  the  tube  and  flask,  we  have 

h'+y=h  or  h'=h—y. 

3°.  Let.  A  and  B  be  closed  and  C  opened. 

The  pressure  at  C  inward  is  h,  outward  is  h+x,  represent- 
ing by  x,  as  before,  the  vertical  distance  of  the  surface  of  the 
water  in  the  tube  from  the  open  orifice  C.  Therefore  the  re- 
sultant of  the  pressures  at  C  is  h+x—h—x,  and  the  water  will 
flow  from  C  with  a  velocity  due  to  the  height  x.  As  the  sur- 
face of  the  fluid  in  the  tube  descends,  x  diminishes,  and  the 
velocity  of  discharge  also,  till  the  surface  in  the  tube  arrives  at 
F,  at  which  time  x  becomes  equal  to  the  height  of  F  above  C. 
In  this  position  of  the  fluid,  the  equation  h'=h—y  still  obtains, 
y  being  now  the  distance  LF.  But  an  equilibrium  not  being 
established,  air  will  enter  the  vessel  at  F  and  rise  to  ho.  This 
operation  will  continue  till  y=o  and  h'  =  h.  During  this  en- 
trance of  the  air,  and  the  descent  of  ho  to  F,  the  value  of  £  has 
remained  unchanged,  and  therefore  the  velocity  of  discharge 
has  remained  constant.  After  this  the  value  of  a;  will  diminish, 
and  the  velocity  of  discharge  also,  till  the  water  in  the  flask  de- 
scends to  the  level  of  C. 

BRAMAH'S    HYDROSTATIC    TRESS. 
434.  In  the  annexed  section  of  this  instrument  L  and  H  are 
vertical  cylindrical  cavities,  in  a  solid  mass  of  metal  or  other 


BRAMAHS     HYDROSTATIC     PRESS. 


297 


strong  material.  The 
diameter  of  H  is  con- 
siderably less  than  that 
of  L,  and  they  commu- 
nicate through  an  aper- 
ture N,  in  which  is  a 
valve  opening  into  L. 
A  is  a  strong  piston  or 
solid  cylinder  of  iron 
fitting  closely  to  the 
surface  of  the  hollow 
cylinder,  and  movable 
m.  it.  IH  is  a  piston 
similarly  applied  to  the 
other  cavity  H,  and 
movable  by  means  of  a  lever  KP,  whose  fulcrum  is  at  P.  At 
O  is  a  valve  opening  upward,  and  below  it  the  cylinder  H  is 
continued  downward  to -a  reservoir  of  water.  The  lever  KI 
being  raised,  the  water  ascends,  as  in  the  common  pump,  from 
the  reservoir  G  into  the  cavity  H.  The  lever  being  then  press- 
ed down,  the  valve  O  closes,  and  the  water  is  forced  through 
the  valve  N  into  L,  and,  acting  on  the  piston  A,  communicates 
a  pressure  to  any  substance,  as  books,  included  between  its 
upper  surface  and  a  strong  cross  bar  BC  firmly  connected 
with  the  solid  cylinder  DE.  When  the  water  from  H  is  forced 
into  L,  its  reaction  closes  N,  and  the  same  operation  is  re- 
peated. 

The  pressure  on  the  base  of  A  (134)  is  to  the  force  impress- 
ed by  H  as  the  area  of  the  base  of  A  is  to  that  of  H. 

Let  R  and  r  be  the  radii  of  the  pistons  A  and  H,  L  and  / 
the  longer  and  shorter  arms  of  the  lever  KP,  p  the  power  ap* 
plied  at  K,  and  P  the  resulting  pressure  on  the  base  of  A 
Then  the  force p',  applied  to  the  piston  H,  will  lie 


But  (134), 


PttR2 

pi      -ftp"* 


298     HYDROSTATIC    AND    HYDRAULIC    INSTRUMENTS. 

R2        L  Ra 

*■   r     J     I    r 
This  press  seems  to  present  the  simplest  and  most  effective 
of  all  contrivances  for  increasing  human  power. 

T  R 

If ^=100  lbs.,  -r=10,  and  -=10,  then  P=  100,000  lbs. 


HYDROSTATIC   BELLOWS. 

435.  This  instrument  presents  an  illustration  of  what  is  term- 
ed the  Hydrostatic  Paradox :  that  fluids, 
however  unequal  in  quantity,  may  be  made 
to  equilibrate.  It  consists  of  two  circular 
boarcte,  A  and  B,  firmly  connected  by  a 
cylindrical  coating  of  leather  or  other  flexi- 
ble material.  CM  is  a  tube  communicating 
with  the  lower  portion  of  the  cylinder. 
Water  being  poured  into  the  tube  CM,  the 
/  '  \  boards  A  and  B  will  separate,  B  will  rise, 
/  \       and  a  weight  W,  very  large  compared  with 

■H- ~7   that  of  the  water  in  CM,  may  be  supported 

/  \    on  B.      The  fluid  in  CM  which  counter- 

)  (    poises  W  is  that  above  P,  the  level  of  B. 

^  Let  k  be  the  area  of  a  horizontal  section 

of  the  tube,  K  that  of  a  section  of  the  cyl- 
inder, or  that  of  B,  and  p  the  pressure  of  the  fluid  NP  above 
P  on  k. 

Then  (134)  ^=|. 

If  o  be  the  specific  gravity  of  the  weight  W,  V  its  volume, 
and  NP=/i,  then 

gp,hk_k 
gpV  ~K* 
.      PV        V 

•*■  ^pTk^'k- 

If  V  be  the  whole  volume  of  the  fluid  in  the  instrument,  and 
II  the  height  of  the  cylinder  AB,  then 


DIVING     BELL.  299 

V'=H(K+A)+M=H(K+A)+ffV~ 

V'K-aVk 
•"'  H_  K(K+A)  * 
Let  now  the  quantity  of  fluid  be  increased  by  v,  and  the  cor- 
responding increment  aH  in  the  height  of  W  will  be 
K(V'+v)-aVk    KV'-aYk        v 


aH- 


K(K+k)  K(K+A)      K+A* 


DIVING   BELL. 

436.  The  diving  bell  is  commonly  a  hollow  cylinder  or  par- 
allelepiped, one  end  of  which  is  closed.  It  is  immersed  with 
the  open  end  downward,  weights  being  added  to  sink  and  keep 
it  steady  in  its  descent.  As  the  vessel  descends  the  fluid  con- 
tinually exercises  a  greater  pressure  on  the  contained  air,  con- 
denses it,  and  occupies  a  greater  portion  of  the  vessel.  If  the 
form  be  as  above  supposed,  to  find  the  height  x  of  that  portion 
of  the  bell  free  from  water,  when  the  top  is  sunk  to  the  depth 
H,  let  h  be  the  height  of  a  homogeneous  atmosphere,  a  the 
height  of  the  bell,  p'  the  density  of  the  external  air,  p  thst  of 
the  air  in  the  bell,  and  px  that  of  water.  Then  the  pressure  p, 
on  a  unit  of  surface  of  air  in  the  bell,  will  be 

p=g{hp'  +  (R+x)Pl}. 
But  the  unit  of  pressure  arising  from  the  elasticity  of  the  in?  rn. 
al  air  is  gph. 

.-.  g{hp'  +  Ql+x)pl}=gph. 

The  quantity  of  air  in  the  bell  being  constant,  and  a  horj  na- 
tal section  of  it  the  same  throughout  (155)  and  (156), 

px=p'a,  or  p=p'.-. 
x 

,        TT  h 

.'.  hp'+Hpi+xp1=p'a-, 

or  — k+Hx+x^ — ah, 

Pi  P, 

and  g  being  the  specific  gravity  of  air, 

ohx-\-Hx+x'i=oah. 


a:=-i(H+(j;t)±vri(H+a/*)M-ffiiA. 


300     HYDROSTATIC    AND    HYDRAULIC    INSTRUMENTS. 


B 


Pii 


xMi 


SEA   GAGE. 

437.  The  vessel  AB  is  perforated  with  holes,  and  has  with- 
in it,  fixed  in  a  vertical,  position,  a  glass  tube,  hav- 
ing one  end  closed  and  the  other  immersed  in  a 
cup  of  mercury.  A  is  a  hollow  sphere  whose 
buoyancy  is  sufficient  to  raise  the  instrument 
when  the  weight  W  suspended  at  the  bottom  is 
detached,  'ftie  instrument  is  allowed  to  sink  in 
the  water  whose  depth  is  to  be  determined,  and 
by  a  mechanical  contrivance  the  weight  is  de- 
tached when  it  strikes  the  bottom,  so  that  the  gage 
will  ascend  by  the  buoyancy  of  the  ball.  The 
height  MP  to  which  the  mercury  has  risen  in  the 
j— L-i  tube  is  marked  on  the  interior  of  the  tube,  by  the 
adhesion  of  oil  or  other  viscid  substance  placed  on 
its  surface. 

Let  h  be  the  height  of  the  barometer  at  the  surface,  I  the 
length  of  the  tube  above  the  surface  of  the  mercury  in  the  cup, 
x  the  depth  of  the  water,  and  MP=Aj.     If  h2  be  the  height  of 
a  column  of  mercury  which  would  be  sustained  by  the  elastici 
ty  of  the  air  in  NP,  then  (155) 

h2(l—h1)=hl  or  h2  =  -, — =—. 

Now  the  elasticity  of  the  air  in  NP+  the  weight  of  the  col 
umn  MP=  the  pressure  of  the  atmosphere  on  the  surface  of  th& 
fluid  +  the  weight  of  a  column  of  water  extending  to  the  bottom; 
or,  p  being  the  density  of  mercury,  and  px  that  of  the  water, 
pha+ph1=ph+p1x. 


■  \  x— — (h2+hl—h) 
Pi 


-oh2+o(h1—h) 
hi 

ah 


1 

in  which  o  is  the  ratio  of  the  specific  gravity  of  mercury  to 
-hat  of  the  water  of  the  ocean. 


SIPHON. 


30] 


SIPHON. 

439.  The  siphon  is  an  instrument  for  transferring  fluids 
from  one  vessel  V  to  another  V  in  which 
the  surface  of  the  fluid  is  lower.  It  consists 
of  a  bent  tube  with  one  branch  longer  than 
the  other.  Suppose  this  tube  to  be  filled  with 
the  fluid  from  the  vessel  V,  and  to  have  its 
extremities  immersed  in  the  fluids  in  the  two 
vessels. 

Let  CB=a  be  the  difference  of  level  of  the 
fluids  in  the  vessels,  p1  the  density  of  the  fluid, 
o  the  density  of  the  surrounding  air,  and  k 
the  area  of  a  section  of  the  tube.  The  press- 
ure on  the  surface  of  the  fluid  in  V  without 
the  tube  by  the  atmosphere  will  be  greater 
than  that  on  the  surface  of  the  fluid  in  V  by 
gpak.  This  excess  of  pressure  acting  at  O  upward  will  urge 
the  fluid  from  V  to  V.  At  the  same  time,  the  pressure  exert- 
ed by  the  liquid  in  the  siphon  at  the  level  of  the  fluid  in  V 
will  be  greater  than  that  at  the  level  of  the  fluid  in  V  by  gp1ak. 
This  excess  of  pressure  will  urge  the  fluid  from  V  to  V. 
Hence  the  resultant  pressure  is  equal  to gak(pl—p),  and  in  the 
direction  DAO.  The  fluid  will  therefore  flow  from  V  to  V', 
and,  since  p  is  small,  it  will  move  with  a  force  nearly  equiva- 
lent to  a  column  of  the  fluid  whose  height  is  a  and  base  k.  It 
is  not  necessary  that  the  longer  branch  should  be  immersed  in 
the  fluid  of  V. 

If  h  be  the  height  of  a  column  of  the  fluid  in  equilibrium  with 
the  atmosphere,  and  AB=.r,  then  the  upward  pressure  on  k  in 
the  shorter  branch,  at  the  level  of  the  fluid,  is  that  due  to  h, 
while  the  downward  pressure  is  that  due  to  x.  Therefore  the 
fluid  is  urged  in  the  direction  DAO,  at  A,  by  a  force  represent- 
ed by  h—x.  As  long  as  x<h  the  fluid  will  rise  to  A  and  pass 
into  the  longer  branch. 


302     HYDROSTATIC    AND    HYDRAULIC    INSTRUMENTS. 


THE   COMMON   PUMP. 

440.  AB  and  BC  are  cylinders  connected  together  as  in  the 
figure,  the  former  called  the  body  or 
barrel  of  the  pump,  and  the  latter  the 
suction  pipe.  At  B  is  a  valve  opening 
upward.  M  is  a  piston  accurately  fit 
ting  the  barrel,  and  movable  by  means 
of  a  rod  EV  and  a  lever  EF.  In  the 
piston  is  a  valve  V,  which  opens  upward. 
The  suction  pipe  BC  is  immersed  in  the 
fluid  to  be  raised. 

To  understand  the  action  of  the  pump, 
conceive  the  piston  M  to  be  depressed 
to  B.  The  air  in  MB  being  thus  con- 
densed, will  open  the  valve  V and  escape. 
If  now  the  piston  be  raised  to  A,  the 
pressure  of  the  external  air  will  keep 
the  valve  V  closed,  the  air  in  BC,  by  its 
elasticity,  will  open  the  valve  V  and  dif- 
fuse itself  uniformly  through  CA.  The 
pressure  on  the  surface  of  the  water  in 
the  tube  at  C  being  thus  diminished,  the 
pressure  of  the  air  on  the  water  without 
the  tube  will  cause  it  to  rise  to  some 
point  P  in  the  tube,  till  the  weight  of 
PC  and  the  elasticity  of  the  air  in  AC  together  shall  produce  a 
pressure  on  a  section  of  the  water  in  the  tube  at  C  equal  to  the 
upward  pressure,  occasioned  by  the  external  air  on  the  water 
of  the  reservoir ;  that  is,  until  the  tension  of  the  air  in  AP,  to- 
gether with  the  weight  of  PC,  equal  one  atmosphere.  The 
valve  V'  being  now  equally  pressed  on  both  sides,  will  close  by 
its  own  weight.  Let  now  the  piston  descend  again  to  B.  The 
air  in  AB  being  condensed,  will  again  open  V  and  escape.  By 
a  second  ascent  of  the  piston  to  A,  the  air  in  BP  will  expand 
through  PA,  and  the  volume  of  water  PC  will  increase  in 
length  till  the  pressure  on  the  section  at  C  shall  again  be  one 
atmosphere.     By  repeated  strokes  of  the  piston  the  water  will 


THE     COMMON     PUMP.  303 

rise  to  B  and  pass  through  V.  When  this  is  effected,  a  por- 
tion will  pass  through  V  at  each  ^scent  of  the  piston,  and  be 
lifted  by  its  ascent  to  the  reservoir,  -above  A,  and  be  discharged 
through  the  spout  G. 

To  determine  the  distance  BP,  and  the  elasticity  of  the  air  in 
BP  after  n  strokes,  let  hn  be  the  height  of  a  column  of  water 
which  will  measure  the  elasticity  of  the  air  in  BP,  and  BP=x„; 
also,  Jet  h„+l  and  xn+l  represent  the  same  quantities  after  the 
(n  +  l)th  stroke.  Let  h  be  the  height  of  a  column  of  water 
equivalent  to  one  atmosphere,  AB=<z  the  length  of  a  stroke  of 
the  piston,  BC=&  the  distance  of  the  lower  valve  from  the  sur- 
face of  the  water,  K  be  a  section  of  the  barrel  AB,  and  k  a  sec- 
tion of  the  suction  pipe  BC.  Then,  since  the  tension  of  the  air 
in  BP,  together  with  the  weight  of  the  column  of  water  PC,  is 
equal  to  one  atmosphere, 

hn+(b—x,)=h  and  h,t+l+(b— x„+1)=h. 
.'.  hn—xn  =  h—b     and  hn+l—x„+l=h—b. 

Now  P  being  the  position  of  the  surface  of  the  water  aiter 
the  7ith  stroke,  and  Q,  after  the  (?i4-l)th,  h„  and  hn+1  measure 
the  densities  of  the  air  in  BP  and  BQ;  and  since,  by  the  (rc+l)th 
stroke,  the  air  in  BP  is  expanded  through  AQ,  (155), 

h,l+1  (Ka + kxn+l)  =  khux„, 
and  substituting  the  value  of  x„+l  =  hn+l-\-b— h,  we  have 
hn+1  (Ka + k  (hll+l  +  &-  h))  =  kh„xn. 

Hence  h,^  =  -\     -^+6-/i}±i\/    — -+b-h\*+4hnxa, 

(  Ka     ,     , .  .        /  (  Ka 
and  fcn+i=f  |  -r+h-b\±\\J  |  j+6-A}'+4^. 

The  length  of  the  pipe  BC  is  necessarily  less  than  //,  other- 
wise the  weight  of  PC,  before  it  becomes  equal  to  BC,  would  be 
equal  to  one  atmosphere,  and,  could  a  vacuum  be  produced  in 
MP,  no  further  ascent  of  the  fluid  would  follow. 

To  find  the  pressure  on  the  piston  when  the  water  is  raised 
to  the  point  N  above  the  piston,  let  it  be  represented  by  a  col- 
umn of  water  whose  height  is  x.  The  pressure  upward  against 
the  base  of  M  is  that  of  a  column  of  wnter  equal  in  height  to 


304.    HYDROSTATIC    AND    HYDRAULIC    INSTRUMENT 


A 


X 


h  —  MC.  The  pressure  on  the  upper  suiface  is  A+NM.  There- 
fore the  resultant  pressure  Vb  the  difference  of  these,  or 
z=A+NM-/*VvMC=NM+MC=NC. 
441.  Sometimes,  by  not  preserving  a  proper  relation  between 
the  distances  AB  and  AC,  the  water  will  cease  to 
rise,  even  though  BC<A.  To  show  this,  let  the 
barrel  and  suction  pipe  be  equal,  or  K==jfc,  and  the 
lower  valve  be  at  some  point  below  B,  as  at  C. 
When  the  water  has  risen  to  X  and  the  piston  is 
raised  to  A,  the  tension  of  the  air  in  AX,  together 
with  the  weight  of  the  column  XC,  is  equal  to  one 
atmosphere.  If  now,  when  the  piston  descends  to 
B,  the  tension  of  the  air  in  BX  does  not  become 
greater  than  that  of  the  external  air,  the  valve  in 
the  piston  will  not  be  opened,  no  air  will  escape 
from  BX,  and  the  water  will  rise  no  further.  To 
find  the  point  X  when  the  air  in  AX,  reduced  to 
the  volume  of  BX,  becomes  equal  in  tension  to 
one  atmosphere,  let  AX=a;,  AB— a,  AC=/.  Then 
BX=.r— a.  Now  when  the  piston  is  raised  to  A, 
the  air  in  BX  being  expanded  through  AX,  the 
h(x— a) 


tension  becomes  (155) 


and 


-(x  —  a)+l—x- 

x  ' 


h, 


x=\l±\JP-4ah. 
If  P<C4ah,  the  value  of  x  is  imaginary,  and  no  point  of  stop- 
page exists.     When  P=4ah,  x=\l,  and.  there  is  one  point  at 
half  the  distance  from  A  to  C  ;  and  when  V>4ah,  there  will  be 
two  points. 

If  a=  10,1  =  36,  and  h=32, 

z=20  and  16. 


AIR     PUMP. 


305 


V 

13 


v 


p' 


-a\£ 


R 


1H 


AIR    TUMP. 

442.  B  and  B'  are  cylinders  or  barrels,  commonly  of  the 
same  size,  in  which  pis- 
tons P  and  P',  with  valves 
V  and  V  opening  upward, 
are  movable  by  means  of 
rack-work,  the  one  ascend- 
ing as  the  other  descends. 
At  the  lower  extremity  of 
the  barrels  there  are  aper- 
tures with  valves  opening 
upward,  and  communica- 
ting by  the  pipe  ac  with  the 
receiver  R,  from  which  the  air  is  to  be  exhausted. 

The  ascent  of  the  piston  P  tends  to  produce  a  vacuum  in  B, 
the  valve  V  being  closed  by  the  pressure  of  the  external  air. 
The  air  in  R,  by  its  elastic  force,  opens  the  valve  v  and  fills 
the  barrel  B.  When  P  descends,  the  elasticity  of  the  air  in  B 
closes  v  and  opens  V,  through  which  the  contents  of  B  escape 
into  the  external  air.  The  action  of  both  pistons  is  mani- 
festly the  same,  and  thus  for  each  descent  of  either  piston  a 
volume  of  air  equal  to  that  of  either  barrel  is  expelled  from  the 
machine. 

Let  V  be  the  volume  of  the  receiver  pipes  and  one  barrel,  V1 
the  capacity  of  either  barrel,  and  p„  the  density  of  the  air  in  the 
machine  after  the  wth  stroke.  Now  the  quantity  of  air  in  the 
machine  after  the  nth.  stroke  is  p„V,  and  by  the  (?i  +  l)th  stroke 
the  volume  Vl5  or  quantity  pnY  lf  is  expelled.  There  will  re- 
main then  in  the  machine  after  the  (?i+  l)th  stroke  the  quan- 
tity p„V— p„Vj,  and  this  being  diffused  through  the  space  V, 
we  have,  by  (155)  and  (156), 

PH-iV=P»V-PrtV1=pn(V-V1), 
V 


or 


P-+i=P»(l' 


-)• 


If  p0  be  the  initial  density  of  the  air,  by  making  n  equal  to 
0,  1,  2,  &c,  successively,  we  have 

U 


306    HYDROSTATIC    AND    HYDRAULIC    INSTRUMENTS. 

Pi=PoO— y-J 

v  \  v\! 

P2=PAl—Y)=Po(l-y1) 

Pn=P0(l—f)- 

It  is  obvious  that  p„  can  never  become  zero  as  long  as  n  is 
finite,  and  therefore,  even  theoretically,  perfect  exhaustion  is 
impossible. 


CONDENSER. 


443.  If  in  the  preceding  figure  the  valves  V,  V,  v,  v',  were 
made  to  open  downward,  it  would  represent  a  condenser.  By 
each  descent  of  the  piston  a  volume  of  air  equal  to  that  of  the 
barrel  will  be  forced  into  R,  and  will  by  its  reaction  close  the 
valves  v  and  v',  and  be  retained  there. 

To  find  the  density  of  the  air  in  the  receiver  after  n  strokes 
of  the  piston,  let  p„  and  pB+1  be  the  densities  of  the  air  in  R  after 
the  nth  and  (n+  l)th  strokes,  V  the  volume  of  the  receiver  R 
and  the  pipe  ac,  V,  that  of  either  barrel,  and  p0  the  density  of 
the  external  air.     Then 

P»+1V=P»V+PoV1. 

•'.    P„+1  =  P„+P0  -y- 

If  n  be  made  successively  equal  to  0,  1,  2,  3,  &c,  we  hav« 

p1-Po(i+:y-) 

V  2V  \ 

P2  =P  i  +P  o~Y=P  o  (1  +_V/ 


CLEPSYDRA.  307 

CLEPSYDRA. 

444.  The  clepsydra,  or  water  clock,  is  a  contrivance  for 
measuring  time  by  the  descent  of  the  surface  of  a  fluid  which 
flows  through  a  small  aperture  in  the  base  of  the  vessel  con- 
taining it. 

Suppose  the  vessel  a  prism.  It  is  required  to  determine 
what  scale  must  be  marked  on  its  side,  that  the  coincidence  of 
the  descending  surface  with  the  successive  lines  of  division 
may  mark  equal  successive  intervals  of  time.  Let  a  be  the 
altitude  of  the  prism,  x  the  distance  of  the  surface  from  the 
base  of  the  vessel  at  the  end  of  the  time  t,  from  the  beginning 
of  the  motion. 

Equation  (172)  gives,  using  x  instead  of  h  —  x, 

kV2ga      k2g  . 
x=a — tA — —t 

Let  &x  and  M  be  corresponding  increments  of  re  and  t;  then 

k  V  2sa ,         .      k*g  , 
x+Ax=a %*-{t+U)+2gi(t+My. 

Subtracting  the  preceding  equation  from  this,  and  we  have 
F°-  (  K     / 2a 

The  time  t  is  in  seconds,  and,  to  determine  the  divisions  cor- 
responding to  successive  minutes  of  time,  put  A£=60",  and  give 
to  t  the  successive  values  0,  60",  120",  &c. 

In  a  vessel  of  the  form  supposed  in  Ex.  4,  Art.  433,  the  vert- 
ical distance  of  the  successive  divisions  will  all  be  equal,  or 
the  surface  will  descend  equal  distances  in  equal  times. 


THE   END. 


CVl 
CO 


